Transcript for:
Lecture Notes on Differentiation

what's the first principle of differentiation so we basically know that if you're given Y is equal to for example 3x uh let's take it to be 3x + 1 you know that when you're dealing with the Dy DX of that function of the derivative is basically going to be equal to three if we add 3x2 and then let's say plus uh plus 2X right so you expect the the derivative in such a case we understand that you need to multiply the power by the coefficient reduce the power by one so 3 * 2 is 6 x reduce the power by one it becomes just one and then all the same X the power 1 1 * 2 is 2 ruce a power by one becomes 1 - 1 becomes 0 x^ 0 is just one okay so you therefore have 6X plus as a solution now according to First principle the first principle is based on the idea that the derivative which we call the derivative function can help you determine the gradient at a given point now if you remember the way you need you used to calate the gradient of a linear function or a straight line if you given two two two what two coordinates of a line right let's say a make up a certain line like that you're able to find the sness of the gradient of a line by saying your gradient as m is a change in y divided by the change in what in X so change in y your Y 2 6us 4 from the other one taking that as a Y2 and this is a y1 this is our X1 and as X2 as well as for the X it will be 3- 2 you get to observe that your solution will be two so what is our gradient now we can use this basic idea to basically differentiate the function that we have they are using what we're calling first principle so basic idea is to take the function to be the y coordinate okay so for the first coordinate we're going to have X y now Y is equal to that so we have 3x2 + 2x as our first coordinate now for the other coordinate which is where the the most important part of this principle is you would have to introduce now um X Plus H so we are saying okay if we have of course using the example of a straight line if this is X or Y plane and then you have a line the basic idea is this okay so we have an original point there now we are now trying to come up with a certain coordinate with another coordinate where we are saying if this is X okay what will happen as we increase by H to the right so that coordinate we are using H to denote the magnitude by which you've moved to the right or the left so let's say it to be there now we understand that for you to find the the y coordinate or the Y value of a function you just need to plug in the value of x now if we taking a value of x to be X Plus H that would mean that we need to substitute in that original part of the function where is X we put X Plus H so going to our three then X+ h² plus now we have 2 x so where is X again put X Plus H the moment you just get that no matter what you're going to have it will all be simple for you so therefore our Dy DX in other terms in some textbooks will give you the formula as the function of x + hus the original function divided by what divided by H so basic idea is this we just trying to focus on the change in y divided by the change in X now note that for the second part the change the Y2 is basically the function of course we know that y represents f ofx so at every part of where we've got the function where we've got X we put X Plus H so that is as good as saying f of x plus h and then f of x is the original part so in other terms all we are doing is this okay we're just subtracting them so subtracting this Y2 so 3x + H sared + 2x + H and then minus the first part which is 3x 2 + 2x and then of course for denominator as well we have x + H - x which will just give us H as a difference okay so at this point we are able to expand so if we expand what is in the brackets there you can agree we have x2+ X2 + 2 HX + h s and then we have if you multiply there we have x + h 2 * x 2 x + 2 H minus if you distribute the negative is there we're going to 3x 2 - 2x okay so to just make sure that we speed up on things here we actually multiply by this three to whatever is inside there so that one become three that one become 3 * 2 BEC six that will will just become three so I can just remove the three there don't forget on the bottom we have what we are dividing by H okay so with that understanding we're able to simplify this so our Dy DX is equal to so we have 3x2 for first part plus 6 HX + 3 H2 and then we have + 2 x + 2 H and then what we subtracting we subtracting - 3x^ 2 and then - 2x so that is able to cancel out with that and then the 3x² is able to subtract that as well and then we're dividing by H which is on the bottom so you can here see that H is common so you can easily factorize it on the top part so if you remove H you remain with 6 x + 3 H + 2 and then the H on the bottom can now divide okay so what is standing out is you have 6 x + 3 H + 2 now for you to find the solution the full answer what you basically get to do is you need to we are dealing with the limits as X approaches sorry as H approaches zero that would mean that if we apply that limit on 6 x + 3 H + 2 it means where there is where is H you put what zero so that will go away so your answer remains 6X + 2 which matches up with the first part there okay so let's proceed and look at a few other practice questions on the same we we look at fractions as well and let's talk about the formula the formula of differentiation so the basic idea that we've looked at so far is we've looked at the first principle of course I understand that you've done this before okay but I still want to break it down so we have 3x 2 + 2x right right so we understand that the Dy DX which of course represent the derivative of that function is going to be 6 x + 2 where have emphasized you need to multiply the power with coefficient reduce a power by one and all that now we need to introduce it now in the form of the formula that's a formula that we using but you need to understand exactly what's going on so what we're trying to say is if one Y is equal to a coefficient of n right let me put it this s if you've got yal a x power n okay if you're trying to differentiate that in respect to X Y in respect to X it has to be multiply the power with a so it will be n a as a product and then X ra the power the power is supposed to reduce by a one so this is the formula for differentiation very important for you to understand okay so that's what we've been using I just needed to mention that but we can look at uh understand that this is easy for you to deal with these polom functions are easy to deal with now let me try to look at rational functions not necessarily rational functions but maybe radical functions and all um let's consider guess what we've got a fraction by the way one over let's say 2x how do you apply that formula so Dy DX before we actually bring that up I need you need you to understand that if you've got two uh maybe I can start with something as simple as 1 /x so if you've got one /x if if you take it to the numerator that's as good as saying X so on the bottom there is the power one if it goes to the numerator it becomes the power1 that's very important so at that point you realize or you understand that if you need to apply the power rule what you do is this folders if okay so this is going to be your Dy DX is going to be equal to the coefficient there is one so1 * 1 will give US1 and then X you need to reduce it by one the power is supposed to reduce by one so it becomes x^ -2 which you can grad take on the bottom by becoming -1/ x^ 2 that's as good as that so that's what you have applying the power rule so all the same even if you had the power two and all if it goes to the top part it become -2 and then becomes -2X and then the power reduces by one becom3 you can take it on the bottom if you want so that you have -2X power 3 okay so now what changes if we are to introduce let's say um A2 there so this is now 1 / 2 X2 so obiously same the fraction will remain the number remain on the bottom but you need to take the X back to the numerator that becomes a negative power now the coefficient in this case becomes half so 2 m half and then X redu it by the power is supposed to reduc by one so it be2 - 1 so this and that will divide you have A1 and then you have x^3 so which is as good ASX you take X from the bottom so be1 /x power 3 okay so that's the interesting part about this so that's basically how you can apply the the power rule which we are coding the formula to just basically perform basic different der basic differentiation okay so as the name in tells product through this is where we talk about uh the product of two functions okay so let's say in the case where you got your y it's a product of two functions let's say you have x + 2 and then multiply by X squ um maybe let me put it this way 2x uh yeah 2x^ 2 plus a one so if you look at this from that point of course this is easy because you're still able to multiply them and all that but the basic idea is you can use a product through in a simpler way to avoid using the power because you take a lot of your time so when you've got product of functions we use a product rout to basically help us to determine the what derivative so the way it works is as follows if you have your dydx you want to dety DX if you've got a product of of two functions okay so you're going to look at it from this way it you can determine the derivative of a first part which in this case you know what if you differentiate what is in the brackets you're going to just have a one to make it interesting I can put a 4X there so if you differentiate that you're just going to get a four multiply by the original part of the other side so we have 2x^2 main 10 + 1 and then all the same we'll do the opposite this time around we differentiate the other part okay so to make it interesting again I'll remove that so I put a three so that you can see what's going on here so if we differentiate what is is on in in that other bracket you going to get 6 x open the brackets we need to multiply by the original part of the other part which is 4x + 2 so this becomes if we try to open this bracket 4 * what is inside there you're going to have 8 x^2 + 4 + 4 6 * 4 is 24 x 2 6 x * 2 becomes 12 of x if we are to collect the like terms 24 + 8 basically becomes 32 and then we have a 12 X plus a 4 so this becomes the derivative now going back to the initial step which is very important what we have is first of all getting the derivative of the first part multiplying by the original part of the other part which now of course I've made a mistake because of the collection I made there so this changed to three by the way so that is supposed to become a three so that if you multiply it becomes 12 now 12 + 24 becomes 56 okay yeah all the same so taking this to be U taking that to be V the product true is as simple as the derivative when you've got a product of two functions just get the derivative of the first part which will call U Prime multiply by the other one V and then add now if you look at the 6X 6X is as a result of differentiating V so take that as V Prime and then multiply by the original other side so you understand that addition is commutative it doesn't matter what you start with so even if you start with a derivative of the other one it is still the same okay you don't have to stress about it but basic this is a basic idea about the the product true for your practice feel free to just to just pause the video and just try out this set the power one this is derory so simple this is derat so simple you can even deal but I want you to apply the product instead of working with the power so all the same Dy DX just to help you remember the formula it's there are no restrictions addition is basically commutative okay so but I like it this way without using the formula the product tells me I just need to differentiate the first part so 2 * 3 will give me six so I have 6 x I need to multiply this by the original other part as it is and then plus so I need now to differentiate this other part I don't know why I'm coming up with such equations which are giving me the same result okay it would be different any 3 * 2 be 6 now the power reduced by one it becomes x s we multiply by the original other part which is 3x2 that's the basic idea okay so I have 6X let me just multiply 6 x multiply what is there 6 * 2 is 12 x * x^ 3 x power 4 + 1 * 6 x is 6 x and then the other part is 18 3 * that 6 by 3 and then x 2 * x² is x^ 4 18 + 12 that's 30 plus 6X so that's how as simple as this is the way it will come out Simply okay so later on after you've looked at the exponential functions it's going to become more interesting because you can actually have a product of the quadratic function like 2x M exponential the power to X that's also a product you can also have logs as well the same concept will still apply okay you have log x² multip by certain other function like an exponential and all so you have to apply the same principle mly the original part by the derivative of the other okay I'll give you an example of this one just in case you're curious so I'll differentiate the first part the derivative of 2x is just two and then I multiply by the original other part and then this other one now I need to start by thetive of e^ X so the derivative of an exponential function is differentiate what the power is there okay differentiate what the power is so the power in this case if you differentiate 2x say two so multiply that so it will be 2 e^ 2x the power doesn't change for an exponential function so our derivative is 2 E power 2 x that's why if you've got E power x the derivative is still E power x because the Dera of X is a one one multip by that nothing will change okay and then multiply by the original first part okay so you have 2 e^ 2x plus and then here if you multiply you've got 4X e^ 2x you can there as a solution but most of the times I prefer to factorize e the part 2 x which is common and then remain in the bracket the first part remain two the other part 4X so again two is still common you can still factorize it out so you basically remain with 2 e^ 2x is common there remain 1+ 2x basically that's what you have okay so this is what we're getting now the product through we done let's move and look at U the question through okay so um talking about the question through just as we've talked about the product through this is what deals with where you've got question right where you're dividing so gue where you've got Y is = to U over V okay so here now the formula tells us that if you want to determine y prime or d y DX which are the same the formula is as follows you need to start with the denominator multiply by the derivative of the numerator minus the opposite which of course we expect to be u v Prime / by v² this one is very important it's very important one that I myself remember is whatever I have on the denominator should come first and then M the derivative of the other and then subtract with the opposite and then divide by the denominator squared that's easy to remember right okay let's try to look at an example so in a case where you have y equal to x² over x + 2 you want to differentiate that so I've said it's as simple as starting with what the denominator the denominator in this Cas is is x + 2 so we have x + 2 right multiply by the derivative of the other part which of course is the numerator so that's going to give us 2x minus the opposite so don't forget we started with the denominator so we're going to start with the numerator which is x² MTI by the Dera what is on the denominator which is a one x + 2 differentiated is a one over the denominator squared this is equal to if you need to multiply that don't forget 2x is multiplying whatever is there so you're going to have 2x2 for the first part and then 4X and then Min x 2 over x + 2 2 so so this can subtract 2x2 - X2 is just X2 + 4 x over x + 2 s on the top part if you want you can factorize so you have what x and then x + 4 so it's not worth it we can leave it there okay so basically that's a basic idea on what you basically are required to do on how to apply this others would prefer to remember the formula and or and then start finding U Prime but I feel it's basically more direct where you think of it is I'll start with the denominator multiply by the derivative of the numerator okay and then subtract the opposite divided by the denominator squared that's a basic quotient rule that you need to remember no okay um feel free to pause the video and just try out this one where you have your y being equal to 2x + 1 and then divide it by 3x + 1 that's fun it's very simple okay so I'll now apply the formula just to make sure that you can if people are interested they can also take note of it so looking at what I've written there my U is what is on top so my U therefore becomes equal to 2x + 1 the U Prime is the derivative so it's just two and then for v v is = 3x + 1 V Prime becomes three so I'm getting these after differentiating so your dyx now becomes according to the formula I said start with a denominator so your Dy DX is therefore equal to your V which is a denominator MTI by the numerator derivative minus opposite which is u v Prime over v^ 2 which is the denominator of course so our V is 3x + 1 and then our derivative of U is a two minus 4 U we have 2X + 1 now you have to be very careful the brackets our V Prime is a three divid by v² which is 3x + 1 the denom S so you're able to simplify multiply the two if whatever is there the then subtract so we have 6 x for first part plus two and then we have minus we have got a three there so I open the bracket 3 * 2x that is 6 x + 3 over 3x + 1 2 of course the top part you expect 6x - 6 x will give us zero and then you have got 2 - 3 which is A1 over 3x + 1 squ that's how simple this is all right so bic is it about the question true let's move on and talk about the chain rule uh let me introduce the basic idea of chain R okay so chain R basically focuses on the idea where you've got I really like thinking of it in the case where you've got like brackets let's say Y is equal to and then you have 2x² plus one the power four in this case chain R is more like a combination of power rule with something new or extra you understand why it's called a chain rle so you take your whatever is inside the brackets to be your U and then the power four okay so the way we basically get to under of this is um the U that you have in this case our = 2x 2+ 1 so that can be differentiated so differentiating that it's D DX because we're differentiating u in respect to X we're going to have what 4x and then we can also differentiate the part where we' substituted U so it's Y in respect to U so it's D U and then we have got a 4 U and then the redu by one to power three okay now you've got this and you've got that so if you try to multiply the two there's one thing that is going to be interesting d y do you multi by D DX so the D will cancel out you remain with Dy DX so in other terms if you multiply the two you're going to get an answer so 4 U the power 3 by 4X you're going to have 16 and then X and then you the part3 now pull back your U your U is what your U is 2x² 2x 2 + 1^ 3 so that's your result and this is why it's called the chain row it's a chain okay that's the basic idea idea using the formula and all that now I like I don't like formulas I don't like formulas so now that you know what we're talking about by the way don't forget the answer that's our answer there on the bottom using what we're calling the chain so the way you can go about directory is as follows consider the coefficient that you have there so this is a case where you only have a coefficient but you don't have another function if you do have let's say where one if you maybe 2 and then m whatever is there the formula changes doesn't this is now it becomes the product of two different functions making up its function so there you have to apply what we're calling the product through so we view it review it don't forget about it for now going back to our question so the form normal differentiation using the power rle you take one to be a coefficient 4 * 1 is four and then maintain what is in the bracket and then reduce the power by one which is a three now and then multiply by the derivative of what is in the bracket that's the only that you add multiply by the derivative of what is in the bracket the derivative of 2x2 + 1 becomes 4X and then if you multiply 4X by four outside becomes 16x now 2x2 + 1 remains in the brackets so you've seen that it's not as complicated as the formula the formula takes and Wast a lot of time okay so you can actually apply the same idea the combination of chain Rule and product R to answer the other part but let me try to give you other forms of or other things that would require you to apply the channel so you can have a case where you have the square root of x^ 3 + 6 you can also apply the chain R there to determine the derivative of such a function so YX becomes now you understand that the square root is as good is add it the power the power half even if you have got the fourth root you understand that it becomes the power one/ four so it shouldn't be a struggle it shouldn't be intimidated when you see such an expression I don't know how to differentiate such a function so uh I don't know why I've written that but this is still y so now the derivative which is y prime or ydx becomes remember the coefficient there is two one now let's take let's make it more interesting let's assume we had the four there and then we've got the four so 4 * half becomes two and then reduce the power by 1 so half - 1 is just going to be half basic frac fraction now the thing that we adding is where we multiply by the derivative what in Brackets the derivative what brackets in this case we have 3 x^ 3 + 6 so you're just going to be 3x^ 2 this times that gives us 6 x 2 what remaining in the brackets we've got x^ 3 + 6^ half we can name there that's still fine I like taking it to the denominator so have a positive power so x^ 3 + 6 can go the denominator so that is now the power half now the power half is as good as putting a square root now so you look to be more understanding of what you're doing if you express an answer like that so that's where you end now for the sake of your practice I want you to try out this one y is equal to 1 over I know student students when you meet questions that you've never solved you feel like you can't answer it but all the same you know that if you've got something on the denominator part you can just take it to the numerator right and then raise it the sign changes becomes negative so all a sudden there's a one there if you multiply your dyx becomes 3 maintain what is in the brackets reduce the power by one it becomes multiply by the derivative what is in brackets which is 2x3 by that it gives us -6x maintaining what is in the brackets we can directly take it on the bottom so that the power becomes positive that's your answer there uh the other thing that I want to mention is uh there I know we'll talk about this later on in detail how to handle them and deal with them but for now just know that the derivative of sin x Dy DX is just as good as cos cosine of x I want you to know that the derivative of cosine on the other end gives us negative s of X so we look at more complicated stuff later onative of T and all but there's an of this to what you're dealing with the general so so the basic idea behind the derivative of this is you can let whatever is there be your U so Y is equal to U and then your U becomes equal to the function itself s of X now if you sayy you notice that in this case your U just becomes U I think this is wrong I've done it in a wrong way um instead we can say it's this part which becomes U so sign U and then your U becomes equal to X so Dy du is now the derivative of of what of s is cosine so you have cosine of U and then the Der D DX there becomes just equal to one so 1 * cosine of U is just COS of U now your U is equal to what x so that's answer in such a case now why am i showing you this I'm showing you this because it becomes more interesting if we add 2x there if we had 2x so if we had 2x this still maintains it still maintains to be cosign of U but this part becomes = 2x so U DX becomes 2 so if you multiply you have two cosine and then the U is still 2x no okay so that's a basic idea there we'll talk about trigonometric differentiation later on in detail now just to end this how do you under equation where you've the product room being combined copying the question again X2 2x 2 + 1^ 4 so we have a product so I said back maintain the first part as it is or differentiate it if you want but I'll maintain it in this I'll differentiate in this case derivative of X2 is 2X and then the other part should be maintained as it is so maintain 2x2 + 1^ four now for the next part as we add we need to differentiate the other one as well so now how do you differentiate this now you know this is where now the the chain R comes in to differentiate this so I said consider it as it is the coefficient is one so we have four maintain what is in the brackets reduce the power by one it becomes the power three and then multiply by the derivative of what is in the bracket so derivative of 2x2 + 1 is just as good as 4X and then multiply by the original other part which is X squ I hope you're not confused this becomes 2x 2 x^ 2 + 1^ 4 + and then this multiply by that just gives us 16 x maintaining what is in Brackets we have 2x^2 + 1^ 3 now don't forget that this is multiplying by the x s okay so if you multiply with what is in the brackets there the X squ can just be part of the 16 x it becomes the par three but I think that is better off so you can end there but I prefer simplifying it for where you can factor out the 2x since it's common and then the other part that is common is the 2x^2 + 1 now to what power here we've got a power four we've got the power three so we get the smallest power which is three start now in the brackets you remain for the first part 2x has been taken there since it's the power four here you got the power three you with a single 2x 2+ plus what plus a one and then the other part plus you the if you factor out the two two from 16 you eight factor out X you remain X squ okay the inter part has been taken the part three but I think these are like terms as well so 8 + 2 gives us 10 so you remain of 10 X2 plus I think did we have a two or something why am I forgetting what was it oh we the one yeah so that's our final answer this is interesting this is just interesting I like it so that's it about the Chanel uh let's talk about the sum and difference rules of which I just still like uh they're basically okay for me I feel like they're just like part of what we do um but the basic idea is if you have y being equal to the summation of two different functions f ofx plus G of X the derivative of of Y which will call I prime is the same as derivative of the individual function itself in this case f ofx plus the derivative again of of G of X okay so the what this means is the derivative of the entire part is equal to of individual summation of two parts so the basic idea is if you are adding two functions and then you decide to differentiate y as as a whole it's basically the same as differentiating these individual functions and then just getting to add them okay maybe to make more sense as we look at an example so if you have y is equal to 3x^ 4 + 2x^ 3 we understand that y Prime in this case you know how to work work out this by the power so it will be 12 x^ 3 + 6 x^ 2 now what we saying according to the sum rule the derivative is the same as since this can be considered to be these can be considered to be separate functions y Prime is also equal to the derivative of of 3x^ 4 plus the derivative of 2x^ 3 so even if you do them separate and come to add them you still get the same result that's the basic idea behind the the sum and the difference R giving you one more example which maybe make it make more sense but uh in this case you can see the necessity of doing that so if you've got that and equal to cos 2x so this is a function in this case our over function is why but this can be taken to be a separate function that can be taken to be a separate function so some rule or even if you have a negative tells us that you can basically get the Y derivative by first of all take this part differentiate separately take this part differentiate separately and come to add the results you have your derivative that's what the root tells us so if you have a negative as well you can put a negative there so we can get the first part um derivative of uh X x^ 2^ 2 now you do understand that here we've got a product of two different functions you can apply the product through and what's a product true try to remember it okay so the product true we apply addition there so addition is Comm it doesn't matter what you start with in this case I'll start by the derivative of the first part which is 2 X and then I'll multiply by the original other part I'll maintain E power two and then plus and then now I differentiate E power two so let me make it E power 2x to make it interesting so the derivative of e power 2x becomes uh what becomes what does derivative become the derivative of the power which is 2x is just a two so the two multiplying the E part X itself we did talked about U derivative of x exponentials and then what else do we have and then we can have uh let me put this can have something like multiplying now we have to multiply by the original other part which is the first part we started with x s that's the product Ru so we have 2X E power 2 X plus 2 E power 2 x x² okay so that's our first part now you can also get the COS 2x as well and then also differentiate it now you understand that cos 2x is a basic trigonometric derivative so here we apply the chain rule so the derivative of cosine is what is negative sign so sign maintain what is attached to it and then multiply by thetive of this part so derivative of 2x is just a 2 so answer becomes -2 sin 2x okay so to get your final result for y prime it will be the summation of the two answers so we have 2X e^ 2x plus so I can say 2x^ 2 e2x minus 2 sin 2x so that's the idea about the sum and the difference rule it tells you you don't have to worry about sprting them you can sprit them and then come to add their derivatives together to get the over derivative of y now for your practice I want you to look at this and just differentiate this um plus natural log of x and then minus X2 y so you've got three parts that you can sprit into individual functions you know the square root is the same as the power half so after you've learned about the derivatives of of logarithms and then also incit differentiation be able to answer this question right it down you should be able to answer this question after watching the entire video okay let's talk about the derivatives of exponentials so I like beginning with a simple part okay so begin with where we've got the base of e^ X okay a b um so you realize that derivative of that is basically the same okay so don't ask questions for now so derivative of that is that's the power x so the basic idea is if you've got an exponential with a base of e so let's say E power 2x I think this will now help you understand what we're trying to do here so the derivative of e the power 2x is the base itself or maintain the original part itself e the power 2x as it is now multiply by the derivative of the power now if you look at the power the power is 2x what's the Der of 2x you know it's two right so your answer becomes 2 e^ 2x all right so now applying the same root to the first part you understand that the power of X is what is a one so if you multiply one by the power x that's why we are maintaining the same result okay so if you've got derivative you to find derivative of let's say E power X2 feel free to to pause the video and just try it out so what you might expect there is ask yourself again what's the derivative of a power what's the derivative of the power the power is x s derivative of that that's 2x so answer becomes 2x now multiply by The Original Part which is e^ X2 now one thing that is interesting about exponentials is the power does not change okay very important so that it about the where you've got the base of e now what we can now look at is where we've got different bases let's say you have y is equal to 2 The Power let's say x what's the answer there how do you differentiate that so now the basic rule that you need to understand is whenever you're dealing with the derivatives of exponentials Dy DX is given as coming up with a general formula let's say you've got Y is equal to and then your base is a and then ra the power ra the part rest the power x for example in such a case the general or maybe let me put you at the power U right you've got power the basic general rule is maintain first of all what you have power and then we've seen that we also need to multiply by the derivative of the power and then we also need to multiply by the natural log of what the base a okay so what that means is for what we have our y prime or Dy DX is going to be maintain first of all what we have and then multiply by the derivative of x which is just one in this case and then natural log of two so therefore our answer reduces to 2x natural log of 2 okay let's look at some more examples so what if you have a base of four and then rest part to X Plus X um squ what's the answer there feel free to pause the video so y Prime is going to be so for the first part you can see that that's U an exponential right so focusing on this what's it derivative so the derivative is going to be maintain it first of all as it is and then multiply by the derivative of a power which in this case our power is 2x the derivative there is what is a two and then multiply by the natural log of a base so our base is four so natural log of four now plus what's the derivative of x s we understand there that's basic power rule so it will be 2x so if you want to simplify or put an answer in a simpler way I like starting with what just a constant so we can have two and then four the^ 2 x and then natural log of 4 + 2x so that becomes your answer as a derivative now at this point I know most of you are wondering why for where we at different base of v it was a bit different so where we at the base of e the power to X we are actually applying the same concept so remember we say that here the derivative is going to be maintain it as it is multiply by the derivative of the power which is just a two now why are we ignoring the part for the natural log since we are saying supposed to be a natural log of the base so we understand that natural log basically means um in case you didn't know natural log means log with a base of e so if you have got e there then you know that it's equal to one so basically if you've done logs log of the same base and then you've got the same number there that is equal to one so natural log of e this that's an inverse so that is equal to one so it's not necessary for us to show that part when we have a base of e that's why we just trying to ignore it and just saying just multiply by the derivative of the power to the original part of a function to get the derivative of an exponential with a base of e okay so for your practice make sure you're able to answer this question try this one out so 3 x2+ e to e the power 4X plus 2 x the^ 4 so feel free to pause the video and just try this one out uh let's talk about the derivatives of logarithms so I'll begin with the natural log before we look at uh the normal logarithms with different bases okay so what's the Der dtive of natural log of x so the derivative of natural log of x is basically equal to 1 / X okay so why is it so soal formula maybe let me give you another example to just trick you D of D of x what said natural log of X2 what do we expect to be the answer there so the answer is 2x over X2 which can simplify to 2 x so you're trying to get the idea there so the basic idea behind the differentiation of natural log logarithms is that if you have natural log of U the derivative there derivative of that is going to be U Prime ID U itself okay MTI by natural log of e now you understand that we've already established to say the natural L of eal to one why am I saying natur log of it's because I know that natural log is got a base of e there it's a l with a base of E okay so it's very easy so each time you have a natural log you work it out like that now let's try to look at uh a more advanced uh example or let's try to look at more practice questions just ensure that we're able to remember so how about Y is equal to natural log of X2 feel free to just pause the video didn't try that one out so obviously we expect that y Prime as well it will be the derivative of this part which in this case is 2x over x^2 I think we've already talked about this one so it's 2 over 2x okay that's repetition what if we say natural log of 3x^ 2 + 2 so all the same what's the derivative of 3x2 + 2 so the derivative of that is what so it's going to be 6X now over the actual function itself 3x^2 + 2 that's a way you differentiate logarithms I can give you another example what if we say we have natural log of 7 x + 1 so the derivative there is going to be what derivative is going to be seven we maintain the original function 7 x + 1 that's the derivative we can go on to just make it more interesting all about natural log of say sin x what do you expect so y Prime so ask yourself what's the derivative of what is in the brackets the dative of s of X is cosine of x now maintain the actual part which is s of X it's U Prime ID U so our u in this case is equal to S of X and then our U Prime thetive of s of X is cosine of x so it's cosine of x over sin of X now we understand that that is equal to of X so that's the answer so by the way this is coming from trigonometry identities okay so don't worry much about it if you don't know we can go on just before we look at the logarithms there with different bases other than e what of a case where we have natural log of maybe for practice try out where we have tan of X so if you have T of X all the same the derivative is expected to be what so derivative of T of X is SEC s of X over the original which is tan of X that's your answer what about the case where we have a y being equal to the natural log of uh the fourth root of x how do you hand such a question so you know that you can express that in terms of natural log the fourth Ro is the same as x^ 4 now from logs from basic logarithms you understand that if you have log of two and then X power 4 we know that the power of what is attached to the log can become there the coefficient what you have four log 2 of X so the same concept applies to logarithm the which is power of X can become the coefficient so you have 1 and then natural log of x so whenever you're trying to differentiate something with a constant what basically happens is you basically pull out the constant okay so we have 1/ 4 and then derivative of natural log of x so that is going to be 1 4 and then what's a natural log of x now the the derivative of natur log of x we say derivative of x which we know is a one over the function itself so the answer therefore becomes 1 4X as a solution to that question just one more just one more what about a case where we have the entire natural log where you have like the fourth root of the entire natural log of now let's make it X squ just make it interesting what do you do in such a case so what this means is this is equal to the enti natural log or the enti function ra the power four that's what it means so whenever you've got you have any function in Brackets to power this is where you apply the CH R you can't just bring the 4 it's different but we know that the two ver inside can become the coefficient of natural log but the four which is applying the enti function can never become the Coe of the coefficient there okay so we understand from chain route to say you can write whatever is in the brackets to be your U so our U is equal to natural log of X2 okay and then what else do we know now we know that now our y will become if what is in the brackets is U then we're going to have the power for so we can differentiate this two parts so du du DX so what's the derivative of natural log of X2 so we said differentiate this part which is 2x divided by The Original Part which is x² this this reduces to 2 /x and then for the other part you differentiate Dy du that is differentiating y in respect to U so it will be 4 U and the power reduc by one becomes three so we understand that CH rout is you need to multiply these two parts so we have 2x 4 u^ 3 now there is you we can pull back what we had natural log of X2 ra the power now three so you can simplify that by combining 4 * 2 which is8 so you have 8 overx open the bracket snat log of X2 the^ 3 so that becomes Your solution okay now let's uh maybe for we practice try out this one where you have the natural log and then let's say natural log of what yeah the natural log and then we can say third root of X2 just feel free to pause video and just try this one out and see what you're going to get okay so let's look at the basic logs now what about a case where you have a different base where you have log two for example and then X what's the derivative of such a function so y Prime becomes so I said get derivative of of what is attached to the log so in this case the derivative there becomes what so derivative is just one and then I said maintain the function the function is X this part again X and then multiply by natural log of a base so natural log of two so I did emphasize to say on the part where we add the natural log let's natural log of x here we add only one X because we understand that natural log of e because when you go the natural log the base is e so that is equal to this part is equal to one so it's not necessary for us to show it and H that's why even if you have x s on top you have 2X and then on the bottom of x s this part is still equal to a one okay so let's look at more questions what about a case where our Y is equal to log 3 base 3 and then 2 x^2 - 1 what we expect there so our y Prime is expected to be so if you look at what is in the brackets what's the derivative so the derivative that we expect is 4X divided by the entire function itself which is 2x^2 minus one now not only is that the answer you need now to add the natural log of the base which is three so that becomes your answer 4X over 2x^2 - 1 the natural log of three okay consider a case where you have y being equal to log of 8 and then you have x - x^ 6 what do you expect there so y Prime becomes what's the derivative what is in the brackets so we have 1 - 6 x^ 5 after differentiating that using the power rule and then divided by The Original Part itself which is x - x^ 6 now we need to add the natural log of the base which is eight okay I believe this is easy now to remember applying the fact that whenever you have natural log of let's say the same one x - x^ 6 it's just the derivative what is in the brackets which is 1 - 6 x^ 5 divided by The Original Part itself now there you the difference is you have natural log of e which is equal to one so it's not necessary for us to show it so in short the rules are the same the rules that apply to to the log and the logarithms are the same okay what matters is how the final result comes out which is the difference here because natural log of V is equal to one so not necessary okay with the confidence that I have that you've gotten the idea here try out log with b seven and and then s of X what do you expect so here y becomes a y Prime derivative becomes what's the derivative of s of X so it's cosine of x and then divided by maintain the origin of part which is s of X now add natural log of seven the natural log of the base okay consider case where you have yal to what about if the same one I say log s and then I add T of X what do you expect so you expect that a y Prime becomes the derivative of tan of X is x^2 x and then divided by the original which is tan of X now multiply by the base the natural log of the base which is natural of seven okay with all the basic ones that we've looked at what about a case where it becomes more interesting and then you get to have your y being equal to natural log and then in the brackets you also have natural log of x how do you look at that one this is where you have to now it becomes interesting this is why you get to be tested if you've understood what we've been dealing with so what we''ve said is what whatever you have in the brackets is supposed to be differentiated now you understand that if you P out that to b u being equal to the natur log of x you understand that if you differentiate this part the derivative of x is what is one and then maintain the function so one /x so for what you have in the brackets here you have 1 /x now divided by the original part so the original entire part in the brackets is natural log of x okay so we have 1 /x so that one is the same as 1 /x MTI by 1/ natural log of x okay so the answer just becomes one over X natur of X in in in in short whatever is on the denominator of the top part just joins the denominator there so you just have one /x natur log of x with that simple example test yourself more how about the case where you have now natural log and then you have log of 2 x² in the brackets how do you answer that one so all the same first of all you need to differentiate what is in the brackets so taking your U to be equal to what is in the brackets log of two and then x² so a basic rule there is what's the derivative of X2 derivative there is 2x divided by the function itself which is X2 MTI by natural log of the base so at least the X can cancel out so you just remain with over x one will go so you just 2x so we have 2 x natural log of two so you have two over X natur now divided by what so I need to be careful divid by what we need to divide by The Original Part which is log base 2 x² okay now for a part where you've got the natural log the Bas is e so no need of you adding the natural log part so whatever is on the bottom there joins whatever is on the denominator so you have 2 over X natur log of 2 m log of base 2 x^2 okay that's what you have is your final result anything more consider a case where you now have the log Bas two outside and then you have the natur log of 2x inside there so feel free to pause the video and try this one out if you can answer this one then I believe you're very confident of whatever we' talked about so whatever you have first of all differentiate what is in the brackets Dera of natur L of 2x so take it separate what a derivative so derivative of 2x is two and then maintain the function itself on the bottom so the two can divide so I just have one /x right natural log of 2 of X derivative of 2x is 2 and then on the bottom maintain the function so you have one/ X as well in such a case so 1 /x divided by maintain the original part there which is natural log of 2x and then multiply by the natural log of the base natur log of two so your answer becomes 1/ X natural log of 2 natural log of 2x so that's our answer for this case I know most of you are actually way better than me in terms of being able to recall and remember most of this formulas but I just want to talk about the proof yeah because you know some of you able to remember for example the derivative of tan of X you know some of you know it's x² X and then you also know that uh the derivative of C of X you know that it's equal to c² X you know that derivative of SEC of X is also equal to SEC X tan of X and then you also know the derivative of cose of X which even right now I can't remember but of course it's equal to cosec of X of X I can't remember all these I've got issues I can the only one I can remember properly is this part so we are going to look at their proofs from the basic um things that we should know so let us just know that s of X the derivative of s of X is equal to cosine of x so the way I usually remember this I trick my brain by remembering this to say okay sign is a bit normal that it is positive it produces a positive result then of course the opposite is true the derivative of cosine of x becomes what it now becomes negative of X okay so this is where everything starts from and everything builds up from so in a case where they give me negative of X I know that s maintains the sign so it would be negative cosine of x I know if I say negative cosine of x I know that cosine does not maintain the sign so it will become positive s of X but all the same we'll apply and only use the ones written on top there okay let's begin with um the derivative of tan tan of X so derivative of tan of X now we know that tan of X so by the way for you to be watching this video I believe you've got basic introduction to trigonometry so you know how to deal with trigonometric identities here so T of X is the same as sin x over cosine of x so if you've not done the identities watch a video the link is in the description about the identities very important as a basic Foundation okay so if you've got a division of functions we looked at what we call the quent through which I believe you've already done okay so the question through tells us theas the easiest way of remembering the question through is by beginning with the denominator whatever you have on the denominator should start first so therefore if this is our y our Y is equal to tan of X we expect that our y Prime is going to be begin whatever you have on the denominator so have cosine of x now multiply by the derivative of the numerator so the derivative of sin of X we know it's cosine of x now minus that's a questional so don't forget that we started with a denominator so we're going to now start with a numerator s of X and then multiply by the derivative of now the denominator so we like do the opposite so derivative of the denominator in this case our denominator is cosine of x so is going to beid the RO tells us the square of denominator so cos square of X so on the next step y Prime becomes so cosine of x m by cosine of x that is cosine s x cosine s xus negative time negative get a positive so positive sin x and then over cosine s x so some of you already know where this is going right so since we've got a denominator of cosine X this is also cos x we can convert the sin x to cosine X so sin x plus cosine s xal to one I'm not teaching identities that's an identity so if I if I'm to make sin squ the subject this can go the other side so s X becomes 1us cosine X so we can substitute that so we have sin s x I'll remove it and put 1 minus cosine s of x cosine squ of x okay so y Prime becomes what so if you distribute the POS posi so you end up subtracting xus the negative cosine s of X coming from the brackets with a positive one over cosine s of X which is a denominator now this part we subtract so you remain with 1 cosine squ of x 1/ cosine s of X so at that point some of you are already conversent with identities you know where the answer what the answer is so y Prime is equal to 1 cosine s x so I said identities aree a prerequisite to this video so 1 / cosine of x is equal to what so that is equal to SEC of X so if the denom is squared even that will be squared so therefore our y Prime is equal to SEC s of X which is the thing that you people I would remember okay so the derivative of tan of X is equal to SEC squ of X so it's not just like people guess all these are gotten so we've gotten the first part I believe you're now able to remember that so I'll just be taking note of the identities we've come up so the derivative of tan of X we've already found it's equal to S of X let's look at the other parts now let's consider Cent the derivative of content so this is a case where our Y is equal to cent of X now I believe you know your identities so content is the same as what one 1 / t x Now 1 / tan of X is the same as cosine X over s of X of course I've skipped one part you know how to perform all these simple Artic Expressions where if you have 1/ S of X over cosine of x it can go back to that okay one/ fraction is like the reciprocal so you just exchange the denominator of the numerator so that's what we have now we can apply again the Cent true the same C true that we've looked at so apply it in this case take your U to be your cosine and then V there so apply the same thing that we applied to tan you'll be able to get the answer now pause the video and just try that out okay so the basic idea is again start with a denominator so our y Prime becomes our denominator is what so we applying the question True by the way so our denominator is of X now multiply by the derivative of the numerator the derivative of cosine of x is negative sin of X now minus so we started with the denominator so we're going to now subtract the numerator the numerator is cosine of x now since we have started we started with a numerator we can now multiply by the derivative of the denominator now the derivative of the denominator which is of X the derivative is what X divid by we're supposed to divide by what the denominator squ okay it's already coming out so have negative sin s of x minus now this part is going to give us minus what cosine s of X now cosine s of X direct we know it's the same as 1us Sin s x over sin x so where is this taking us soga sin x - 1us so if we distribute the negative we're going to have negative sin s x plus sin s of X and then minus 1 over sin x so I collected the like terms in the Der Tre I know after Distributing this negative this becomes negative and becomes positive so this will go away and then what remains so at this point what is remaining is1 over sin s of X which we now know I believe we know what 1 / s of X is right so 1 / sin of x from your basic identities 1 / sin of X is the same as cose of X okay so since it is squared we maintain the negative that so it's negative now since squared it's going to be negative cc^ s of X okay so and that is true so the derivative can now write that one down as well the derivative of content of X is basically equal to cant s of X trust me I can't remember these I can't memorize them but just I know what I can do to just if I've got enough time in an exam I'm able to just compute the answer and then we can look at the other one let's say case where we have y is equal to Second of X now second of X you know that this the same as one over cosine of x okay that's an identity so this is more straightforward and direct so again we apply the question true now the question true tells us to C Ator so start with COS of x m thetive of the numerator Thea of the constant one is z minus we now start with a numerator which is a one by of denominator now the denominator is co of X thetive of COS of X is what of X now Co Ro tells us you divide by the square of the denominator which is cos square of X this part is a zero and then you have now positive sin of X over cosine s x so cosine s x is the same as cosine x cosine X you know where that is taking us now so y Prime becomes what sin of X over cosine of x is what so I can take this to be a fraction have S of x cosine of x is T of X and then 1 / cosine of x is what 1 / cosine of x is SEC of x this is a part now where it gives us that as a result so the other one that we've obtained is the derivative of SEC of X is equal to SEC of X itself M tan of x okay now feel free to pause the video and try out the other one [Music] so let's say our Y what what is remaining we've dealt with C we've dealt with SEC now remaining with cosecant now Y is equal cose of X now we know that cant is the same as 1 / of X so we can apply the same principle there so what do we expect so y Prime so I've said ready to say the coent root tells us that start with a denominator s of X and then multiply by derivative of one which is a zero as well and then subtract the numerator now multiply by the derivative of the denominator so derivative of s of X is cosine of x and then of sin s x so on the next step I'll continue that this part is a zero so we have negative cosine of x over sin s of X so sin s i can sprit it again I can sprit it solve s of x s of X and then I introduce a one there so cosine divid by S is what cosine divid sign is what so that is content now there negative don't forget and then one / s of X is what cose of X so this is these are some of the things that help some of you to be able to memorize the formulas so the derivative of cosecant of X becomes cant of X itself now there's a negative don't forget so negative cosecant of X um and then content of X I can't easily remember that okay so at this point including the basic ones that you know derivative of sin of X is equal to cosine of x the derivative of cosine of x is sin of X okay so you now know how to derive the derivatives of the basic trigonometric functions tan of xent of X cant of xant of X sin of x cosine of x and these are very useful as you've seen from the most of the questions that we've looked at okay okay so in this tutorial we're just going to understand how we go about implicit differentiation so what exactly is implicit differentiation so I believe we know what differentiation is first of all so if you've been given 2x^ 2 + 4x - 2 so if you ask you to differentiate this in respect to X all you're doing is you're assuming Y is a subject if y becomes a subject then you we are saying you're differentiating to X Dy DX so you multiply the power by the coefficient and then reduce the power by a one so 1 * 4 be a 4 1 - 1 will be a z so of course the derivative of a constant is a zero so that is basically our result now in a case of implicit differentiation where actually trying to differentiate the impit function so what is exactly what exactly is an impit function so if you look at the equation that I've written this is imp why because in this case it is not possible for you to make y the subject of form so it's very very simple so for example I can give you a normal function an excit function where you have 2 y - 4x if I ask you to differentiate this in respect to x what I going to do you know it's possible for you to make X Y the subject right what do you do so 2 Y is going to be 4X and then y becomes 2x so now if you try to differentiate now you find that the itive of of 2x is what the answer is two okay now the idea of impit differentiation is built on this very idea so if you look at the 2 Yus 4X do you know what we can decide to differentiate it I'll demonstrate what imp how we go about the implicit differentiation so we are trying to differentiate the entire equation in respect to X okay so the moment you just differentiate something that has got nothing to do with X you need to introduce d y DX so y for example we we differentiate the normal way Y is power one in that case so 1 * 2 be a 2 and then 1 - 1 so as we say the power reduces by one so you will notice that the Y will basically go away so you just with two now because we are differentiating y we put Dy DX why because we are differentiating respect to X and there in this this part is no X so put dyx and then- 4x thetive of - 4x isus 4al to Z so the idea of implicit differentiation is you need to make dyx subject so therefore if you add four both sides this negative and the positive one will go and then divide by two both sides this two will go away so you end up with the same result as as in making where we had made y the subject right the answer is two so with that understanding we are now capable of simplifying the question that is on top init okay so if you look at the first part we have x s so what is x s if you differentiate that what's the derivative is 2x2 2x sorry the power reduces by one and then y^ 2 the same it will be 2 y the power reduces by one now we are differentiating in respect to X so I'll put Dy DX and that is the only r that you need to remember always add Dy DX in a case where you differentiate a letter that is not of our interest okay and then- 4x the dative of - 4x is what is-4 the dative of 5 Y is what is five now we we've differentiated y so put divide X again that is the only thing that you need to remember very important the derivative of a constant is always zero so we just remain zero on the right hand side now at this point we need to maintain the item that are having y on the left hand side so we have 2 yide x and then we also have + 5 dyx so every other term would have to go the other side right so we are remaining with 2X and - 4 so we have 4 - 2x so we have to now if we look at our left hand side we have to factorize out the dyx since we we want to make it the subject so after we do that we have d x in the brackets you have 2 y + 5 on the right hand side you have 4 - 2x so we know what to do there our goal is to make DDE X subject right so ultimately You' have to divide by this on both sides so you just with what it with it on the other side so that it be 2 y + 5 so that is our derivative of that implicit function and that basically introduces the idea of implicit differentiation of course just one more example to just help you understand one thing that is very necessary very important in terms of the product TR the way we tend to apply the product TR when know dealing with uh with implicit differentiation now consideration where you have uh y^ 2 and then multiply by 2x so in this case how do you differentiate this simplicit so the basic idea is uh this is now going to be considered to be a product true you okay to be product true product true is the one that's going to apply can even add some more functions to just make it more interesting okay so product Ro tells us demands us to to consider first of all any term of your choice I'll start with y^ 2 and then I'll multiply by the derivative of the other so the derivative of 2x is a 2 plus so in the first part I considered y^2 now now consider 2x I'll start with 2X and then I multiply by the derivative of the other so the Dera of the other is is y^ s in this case so y^ s when youate it will be what 2 y now the fact that we're differentiating y instead of X we have to put X that is the only rule whenever you differentiate a ter having y add Dy DX okay so everything else Remains the Same thetive Y is what is 2 y now because it's y we're putting Dy DX because we differentiating respect to X the of 2x is just a normal a two okay so at this point again the same rules apply you need to remain with what the having dyx on the left hand side so I expand this term so you notice that 2x * 2 y will be 4X Y and then we have aide X the other I think Dy DX is 2 y Dy DX everything go the other side so we have this part to be -2 y^2 and then the constant to also go the other side okay so at the point where we've reached now all we can just do is uh the same procedure just factorize Dy DX right on the left hand side so we have Dy DX what What Remain was 4X Y for the first part and the other part will remain with 2 y on the other side we have -2 y^ 2 - 1 so to make things easier we divide by this both sides so that we just remain with 4 x y + 2 y and that basically summarizes the idea of implicit differentiation if you've understood whatever we've done I believe you're okay so make sure you subscribe to have access to more other videos related to math chemistry physics and biology okay let's straight to understand how we get to apply implicit differentiation to exponential functions okay so if you look at this case we have uh this is our function and then of course we want to determine the derivative of course in respect to to X Dy DX y respect to X okay so we know the basic standard of differentiating a function some of the basic things that I would have to remind you is if you're differentiating a term x^ 3 all you just have to do is multiply the power by the coefficient okay so there's a one there 3 by one gives us a three and then the power would have to reduce by one so it becomes two that's the we differentiate the normal functions okay now I have also to remind you to say if you have a constant constants of course becomes zeros and then of course uh let me talk about the exponential so if you have E power x it derivative is what multiply the original function by the derivative of a power so you agree with me that the derivative of x is one so the derivative Remains the Same now the moment I put let's say let me put 2X now I'm saying the derivative of an exponential function is multiply two right so multiply the derivative of the power to the original function so derivative of 2x is 2 so we have 2 e^ 2x so that is how we differentiate the exponential functions now with the understanding of the basic differentiation and of course the exponentials we can straight go go and deal with our question now if you are new to implicit differ I would advise you to watch video okay with a tag on top and also the link is provided in the description you can watch that video on impit differentiation okay this will give you a background if you have 2 x y^ 2 okay plus I can add something else let's say x² and then I say 3 y cubed you know if you get to observe this this case uh you want to differentiate the function respect to x d y DX of course and then you cannot make y the subject we call that implicit you can't make something the subject okay so that's what we're trying to talk about in this case so how do we differentiate it in case where we have an exponential function okay so having an exponential function we are given in this case it is going to be very simple so I know this is easier that is easier I'll focus more on the middle part first of all we try to work out which is e right so we have E power x^ 3 y^ 3 okay so I emphasize to say the derivative of an exponential function with a base e just multiply the derivative of the power to the original function so just have to multiply the derivative of what so the derivative of the power which in this case is X3 y 3 and then again is the function itself which is e x^ 3 y^ 3 right okay so we need to find the derivative of this okay so we have a product of two different things okay so how do we under that so this requires you to understand the product through if you not if you don't know about product you can also watch a video provided in the link the link is in the description okay so product tells us to say that if you have two functions multiplying in this case we have got x the^ 3 and y^ 3 I not like any formula you just go direct so you can get any term to be the first ter to start with in this case I'll start with X power I'll start with Y power three and then I would have to multiply it against the derivative of the other part so what is the derivative of x^ 3 so I explain how to differentiate the functions right it's going to be 3 x^ two the power reduces by one the power multiplies the coefficient okay plus so that's the product now in the first part we had started with y^ 3 this case we're going to start with x^ 3 and then multiply the Der of what the other part so the other part in this case is y power 3 so if you differentiate y power three ignoring the fact that it's a y you get 3 y squared now the fact that is y and by differentiating in respect to X we have to add Dy DX okay so simplifying all this uh what do we remain with so we have 3x^2 y^ 3 plus so there equal we can say we have 3 y^ 2 x^ 3 Dy DX right okay so I can allow me to raise increase just some space okay at this point where we are we've determined derivative of uh this guy which is our power right so we can just quickly substitute them okay so I'll go back to the original function so Dy DX so of course yeah okay there a zero there to zero so what is the dtive of x^ 3 we talked about the derivative is what 3x^ 2 the poweres by one of course the the coefficient one now the middle part I said when differentiating exponential functions you just have to differentiate the power so the Dera of the power is exactly what we have on the bottom there if you differentiate x^ 3 y^ 3 so we have 3x² y^ 3 + 3 y^ 2 x^ 3 Dy DX now you have to multiply everything there multiply by what the original function itself which is e^ x^ 3 y^ 3 that's how you differentiate exponential functions and then the derivative of a constant is just a zero okay ultimate this is what remains now whenever we're dealing with uh implicit differentiation after you perform the differentiation you have to make Dy DX to be the subject of a function or expression that you get to have so remember we have got a function multiplying everything in the braet right so we can basically get to expand so we have 3x2 plus 3x^ 2 y^ 3 which is like the first term mtip by E which is now coming from outside and then again plus 3 y^ 2 x cubed time YX and then now also time e x^ 3 y^ 3 = to Z so our goal is to make the part X subject so everything else goes the other side so remain with 3 y^ 2 x cubed Dy DX and then we have e^ x^ 3 y^ 3 is equal to so these two terms are going to the other side they'll be become negative right so we have 3x 2 - 3x^ 2 y cubed exponential x^ 3 y^ 3 okay so at this other point where we've moved now it becomes easy to simplify it because we've almost made Dy EX object so all we just have to do is divide by this part and that part okay so after doing that what do we expect to have so we can say our dyx is going to be the left part which is - 3x^ 2 - 3x^ 2 y Cub e x^ 3 y the^ 3 and then we are dividing by what this 3 y^ 2 x cub and then exponential x^ 3 y^ 3 okay so at this very Point we've made Dy DX to be the subject of formula so we've differentiated we've differentiated using the implicit differentiation and then just one more question to make sure that everything of done are stuck okay so this is a second practice question so we have 5 y^ 2 + 2x + 8^ 3x y how do we get to differentiate this okay so in prit differentiation I'll go direct 5 y^ 2 that will give us 10 2 * 5 is 10 the power reduces by one it will be y^ one so the fact that we're differentiating respect to X Y in respect to X would have to mply by D YX plus so 2x this is just we're differentiating X which we're interested in to just be a two and then the other part where we have exponential so I said when you're dealing with exponential you have to first of all determine the derivative of the power in this case the power is 3xy so we have products of two functions okay we can say we have 3x and also y so what is the derivative of 3x it is a three and then what is the derivative of oh we have to M by the other part which is y Plus what's the derivative of y so dative of Y power one is just going to give us 1 * dyx why am I saying M DX because we differentiating y so it's d y DX and then we multiply by the have a part which is 3x right it's that simple so this is product true so the derivative that we have gotten we supposed to multiply it against the original exponential function so the original exponential function is E power 3x y so at this point where we have reached we are now at Liberty of trying to make a dyx before we could get to do that we have to expand the other term so we have 10 y Dy DX uh + 2 so we have 3 y multiplying by the exponential function to be 3 y so of our exponential function and then this part as well 3x multiply by the exponential and then of course not forgetting our dydx equal to zero so if you try to look at what we have how many parts are having the expon the dyx those are the two parts right so those ones are going to on the left hand side so obviously we have I can directly factorize YX with much of your time so divide X and then Remain the first term is 10 y and then then the other one is 3x uh exponential 3x y equal to what is going the other side just this two terms so have minus 2 minus 3 y e 3x power y or E power 3x y so at the point where we've reached um we can just just divide both sides so automat we're just going to remain with uh 10 y + 3x e^ 3x y so that is basically how simple it is to determine the the derivative of exponentials using implicit differentiation okay so just like I said if you want to understand more how we go about it differentiation check out the video in the description and also understand more about the product it will make things easier how you get to deal with this concept so thank you very much for taking some time to watch this video um The Journey under calculus is is becoming more interesting and interesting so we've already talked about differential calculus let's try to talk about the tangents and the normal lines so I'll I actually explain this and then just look at two questions to help us solidify whatever we're going to to talk about considering this Cave of ours taking that to be our X Y plane okay so considering a thin point on the cave so there are two lines there there's one line that will cut the cave at a single point and then there will be a line which will be perpendicular to to that point so the line perpendicular is called the normal and the line parallel is called the tangent now coordinate now coordinate geometry is very important so when you talked about the linear linear equations you understand that when two lines are parallel it means their sness is the same in other terms the gradient of two parall lines are equal and then for the perpendicular ones so that's if I introduce a line perpendicular meaning that they make an angle of 90° and taking that to be M3 their product M1 and M3 is equal to A1 that's very important so with that understanding we understand that the gradient of a cave at a given point is going to be equal to the the the gradient of a tangent so taking the gradient of the cave at that point to be M and then that of a tangent to be M1 that is a relationship there and then for the normal and the cave their product is a negative in other terms m is going to be1 over the gradient of the normal so that relationship is also very important and I think that's what all you need to know but the other thing under calculus is the derivative that's if you have for example an equation Y = x^2 and then you try to determine the gradient function of the derivative your dydx is going to be equal to 2x so you understand from the introduction where we talked about first principle I did emphasize to say this is called the gradient function meaning that it's can help you find the gradient of a cave or any function at any given point so given the value of x at a certain point I'm able to find the gradient okay so now be using this relationship now to help us find the equation of the tangent and the normal to 10 given point just using the derivative and then the x coordinate given okay so let's look at an example to just understand what I'm talking about considering a case where our Y is equal to X2 that's our equation the coordinate to be 3A 9 okay so we need to find let's try to find the tangent to v c at this given point tangent and the normal so I'll find for the tangent and then you find for the normal so let me try to show you how so the first thing that we want to do is first of all try to find the dydx derivative is 2x Okay so so that is called the gradient function now we can find the actual gr gradient at the given point in this case our value of x is three so if you substitute our three we'll get six so this is our gradient of a cave at a given point so of course uh it doesn't matter whether you know how to sketch the function or not but I believe this one easy to sketch this this is a function that's like this it doesn't cut it doesn't go to negative part like that so taking 3 comma 9 is somewhere here 3 and then comma 9 so we're trying to find this tangent and then also the normal so starting with a tangent a tangent is a line that cuts a cave at a single point in this case the point is 3A 9 now what did we say now I did say that the tangent and the cave at a given point their gradients are equal so the gradient that we've determined which is over a cave is basically also equal to the gradient of a tangent at that very point so meaning that we have our gradient to be six now tangents and normals are lines so if you have a gradient and then you have the coord 3 9 you should be able to find the equation how so you know of this I believe why is MX plus which is the form of a standard linear equation so plugging into you find the constant this is our y our Y is 9 our gradient is 6 our X is 3 so 9 and then 6 * 3 is like8 which is equal to C so C is equal to9 so you found you C so the equation becomes y equal to the gradient is 6 so have 6 x and then you C is 9 so- 9 so this is the equation of the tangent to the C at given point okay so how would you handle for how would you handle it for the normal so for the normal the product of the gradient of the normal line of the normal to the cave at a given point the product of its gradient with the gradient of the cave which in this case is six their product is equal to1 meaning that for our normal is going to be1 over the given gradient and that is always a case it's ne1 given the the the gradient of the normal is1 over the gradient of a curve so you found this now we have our coordinate to be 3A 9 that's a point where the normal cuts the cave so using this and applying the same principle find the constant C and then just write the equation simple to deal with right now to just make it more interesting look at a more advanced question which is very nice okay so pay particular attention to this equation to this question that is the KFC so we have a KFC take note as an equation that so write the equation the equation is 2x Cub plus x^ 2 - 3x - 7 that's our equation now the point p with the x coordinate 0.5 so if if I were a you I need to find the y coordinate in advance because I know it's going to HT me as I proceed so how do you find the y coordinate the y coordinate uses the equation of the cave you know that it is equal Y is equal to that so plug in the value of 0.5 at every Point okay so after plugging in the value I'm getting here is8 hopefully my calculations are okay okay so we have that point P now we read the question again the cave c as that equation the point p with the x coordinate 0.5 lies on C so to just help you understand I can come up with a simple sketch we can take that as our cave and then we can have a point so we have a certain Point 0.5 so let's say somewhere there let's take that to be our coordinate 0.5 0.5 comma 8 which we've already found so now this is a point P so the point p with x coordinate lies on C so this cave is called C and then what else are we told the tangent to C so the tent to C so have a tent at that point at p is par so that's this tangent so that's the tangent to the K of C at the point P that tangent is now parallel to a straight line so there's another straight line okay I can just put it down so this line is parallel to the tangent now this line the equation has been given to be what the equation has been given to be 2 y - 3x + 4 = 0 okay so find the equation of the tangent at p in the form that so we just need to find the equation of the tangent now the beauty in this case is if you look at what we have it's somehow it's somehow simple and straightforward I don't see the point I don't see the reason why they gave us this tangent though I don't see the reason maybe we understand because I feel the information given about the cave is important is necessary and sufficient to find let's try to look at it so the cave C the equation has been given to be this part okay so I believe we're able to find the the gradient from there and then the coordinate is already there so I don't see the point anyway let's try it out so Dy DX becomes if you differentiate this you're going to have 8 x^ 2 + 2x - 3 now we can plug in the the negative 0.5 which I believe is uh thetive 0.5 am I even was I even correct there I think there something I made a mistake somewhere when finding the coordinate of the y coordinate so anyway the value of x is 0.5 according to the question so 0.5 is the same as half so 8 x put half so half s is 1 4 and then here we have a half again - 3 so there we're going to have a 2 - 1 - 3 2 - 1 is what is a one - 3 -2 so answer is -2 so that is the gradient of that cave that point so M isal to -2 so we can just find the y coordinate correct to help us find the find what to help us find the the equation of a tangent anty what else do we need to know so let me try to riew finding the the y coordinate there so we have two now whenever you raise a negative number to an old Power it will be negative so going to1 8 and then when you raise 1/ 2 power 2 it will be POS 1 / 4 and then - 3 1 / 2 - 7 so at this point we have 1 over 4 for first part + 1 / 4 which I believe is a zero and then plus 3 / 2 - 7 so if we do multiply there 3 2 * 7 is 14 and then you have got a two on the denominator 3 - 14 is1 I believe so1 / 2 what is our y coordinate I believe if everything is correct there so1 / 2 so at this point I think I can leave you the gradient is -2 the coordinate has been given you able to use this format to find the so this is why I was saying I don't I didn't see the point of them giving us the line parallel to that but let me try to look at it if it's going to add up so let's try to make y the subject in this case so 2 y becomes = to 3x - 4 and then divide by two throughout y becomes 3 / 2 x - 2 so our gradient according to this equation is supposed to be 3 over2 so if you're basing your calculations on the line par to it this is supposed to be the gradient so I expected them to be the same because I'll read the question again the c c has equation that the point p with x coordinate lies on C okay and the tangent to C at p is parallel to a straight line that find the equation of the tangent being the form of that so anyway based on the question but in the actual sense first I saying it wasn't necessary but I feel they wanted us to focus on the usage of the parallel line to help us find the equation so all we can basically do is the thing that is very important is to not that the gradient of that line is equal to the gradient of the line which is the line to which it is parallel to I think that's the most important point the take your point so if in a case where the the what where the equation is not given if equation is not given you should be able to Al the formula in the form of y mx plus C and get M as a gradient of this other line and then use it for the other line but I feel what they needed to give you again is they would have to give you the y coordinate okay so in this case take your gradient to be 3 over two which is coming from the line parallel to the tangent and then get the coordinate 0.51 /2 and try to find the equation okay so this is where we end for this tutorial I hope you understand how to find the normals and also the tangents to the cave and being able to apply them in different questions that may come as complicated let's talk about the concept of uh increasing and decreasing functions so this is under differential calculus okay so we already know how to determine the derivative of a function so the derivative of a function can tell whether a function is increasing or decreasing but before we look at how let's try to understand what it means for a function to be increasing or decreasing okay so consider a linear function like this um and then consider another linear function from the other direction like that so which one would you say is increasing and which one would you say is increasing so we basically get to observe that by looking at the x axis so if you look at our xaxis let's take this one to be maybe somewhere ne4 and then this side positive4 so the question that you need to ask yourself as as we are moving from the negative xais towards the positive what is happening to the Y value so obviously you can actually note that for this other one as we move towards the right the value in y is what we can observe or not an increase in the value of y in other terms the function is increasing but for the other one where we the value of y is kind of decreasing we can see that there's a decrease in the function value okay so this is basically what we're trying to say now of course if you try to look at the equations of these linear functions it's it's an undeniable fact that if you try to determine the equation likely the equation is supposed to be in that form for what other one and then this one as well okay so explicitly mentioning that for this other one m is of a sness is expected to be negative the other one positive okay now we've also talked about the derivatives to say the derivative of a function can help you determine what the gradient we call it derivative function or the gradient function okay so I think you've now gotten the point there so you know what it means for a function to be increasing and you now know what it means for a fun function to be decreasing now of course calculus is more concerned or interested in talking about caves that make it more interesting so let's look at the caves and just basically see what we might expect okay s that so if you talk about an exp exponential function we know that an exponential function behaves in this way like that okay so you can see that as we move from the negative xaxis towards the positive part the value of y is doing what is increasing so that's an example of an increasing function so that is of course uh power x now of course we can also have another example of uh the logarithm function as well something like this so that is also another function you can also see now that it's also what the value of y is increasing as we move from the negative or from left to right we can see an increase in the value of y so that is also an increasing function okay so examples of decreasing functions would be the ones where it's like going down as you move from the left to the right okay it can be like this as well the value of y is reducing As you move from left to right that's what differentiates decreasing and increasing functions now these are perfect Kind of Perfect functions that we've talked about so far now there are also some functions where it's mixed up depending on the Range that you're talking about the function behaves differently at different domains so you can see clearly to say if we try to observe this function at this range at this domain we've seen that it's increasing by that line here you can see that it's decreasing continues to decrease and then again starts to increase so it depends on the kind of functions that you're talking about so the domain also does matter so now this is where we're going to be working with uh the critical points and stationary points to help us determine for certain functions where it is increasing and where it is decreasing but of course we've understood what it means for a function to increase or decrease so I think this is going to be more interesting as we talk about cave sketching that's why understand to say okay a function can be increasing decreasing and be constant at given points okay so theoretically talking before we even add to C sketching let's look at a few functions and try to check whether they are increasing or decreasing functions in general uh let's consider a function y is equal to X2 so this is a very common function so even before we apply calculus let's try to sketch it so obviously you know that this is squared so you don't expect to get any negative at any given any value of x so this is the graph is going to be okay so of course one statement that you can notice um you can clearly see that as we move from left to the right before zero what are we saying the function the value of y has reduced so considering that domain so from the negative Infinity all the way up to zero we can basically see that the function is what the function is decreasing and then after this what can we see so from zero again in increasing so from zero all the way up to positive Infinity we can conclude to say function is now increasing okay so with Calculus what you need to understand is when the derivative or Dy DX is greater than zero when it's greater than zero or when you have a positive value in other terms then you need to conclude that the function is increasing in a case where it is less than zero in such a case you conclude that what the function is decreasing on that interval okay so of course there are some cases where the DDE X will just be equal to zero but we talk about about that for now take it as greater than or equal to Z to be an increasing function okay a case where it is equal to zero will make sensibility okay let's move on to a different kind of uh a function where we have let's change the par let's make it the par three so this one is a bit unique compared to since this one can accommodate both positive values so when you're on the positive part values will continue to be positive then as you go to negatives it will be negative values alone so this function behaves like this behaves like this and then again behaves like this so I actually didn't finish on the Y = X2 so yal x² if you differentiate it or determine the derivative our DX is equal to 2x so if you consider the part of course the graph is something like this okay I hope you've not forgotten for X2 the graph is never cuts the X the negative part of y axis okay so if you consider values less than zero going the side you can get a negative one you plug in the dydx so if you plug in there what do you expect you plug in a negative one the value will be-2 so a negative value remember I said when the Dy DX is less than zero then the function is decreasing so we can c that is decreasing of course if you get the positive part you expect that it will be give you as a positive value which indicates that the function was increasing now of course I can now mention that if ayx is equal to Z that is referred to as a critical point or stationary point you understand what stationary points are so stationary because this the function is nether increasing or decreasing at that given point okay so you understand so y x^ 3 if you look at what I've drawn here what you might expect is after zero going this side positive from zero to positive numbers we can see that the function is increasing as we move from the left again we can still see that the value of y is still what is still increasing so how do we tell how do we know so let's try to differentiate this function so according to what we've looked at we've seen that the function is purely increasing throughout so dyx hopefully to confirm that if you differentiate this function we going to 3x2 now now this is where it becomes interesting whenever you've got X or any valuable raised to a positive even number or let me just say positive power not necessarily positive power but an even power 2 4 6 8 and so on and so forth in such a case you can never get a negative not unless if you have a negative coefficient but in this case it is just like that 3x2 so the values that you're going to plug in plug in a negative where is X they'll continue giving you positive values so even if you get the values less than zero negative what you raise the power to it become a positive so here all values will always be positive which clearly proves that this function will be increasing throughout okay I think that's very Cardinal and very important to just help you understand okay so with linear functions of course it becomes very easy because you know you can just write it in the form of yal MX plus C so if you make y the subject here you would have to divide by four through out so Y is equal to of course take the other part to the other side and then you have plus 7 so you can clearly see to say the gradient M there is a negative so this is expected to be a linear function that is what decreasing even if you try to differentiate this part y prime or Y DX is just going to be3 over 4 so a negative value indicates the function e is decreasing okay what about if you have certain interesting Parts what if you had F ofx and then rest upon Negative X I'm so curious to just try to understand how you would look at this would you say this function is an increasing function a decreasing function or what would you say so let's try to determine the derivative so you can not easily sketch this function so let's try to determine derivative now of course we understand we apply the we have to apply the product rle we've got two different functions here so I'll take the first one X I'll differentiate it I'll get a one and then I'll multiply by the original other part and then I'll add I'll start now by the derivative of e Negative X so we said how do you differentiate exponentials so derivative of an exponential differentiate the power so power there if you differentiate you get a negative one so now negative 1 multi by by the original itself that's the derivative of an exponential function okay of course you have to add the natural log of natural log of what natural log of the base itself which in this case we just cancel out and then multiply by now the original first part so at this point you can factorize e^ Negative X which I believe is common for the first part remain with a one the other part you remain with a negative and then X this is the derivative that you have okay so one thing that you know is an exponential function if you try to sketch it it never cuts the x axis this is the way it behaves it will always be getting closer and closer to X as you move towards the negative numbers and we increase exponentially like that so me meaning that no matter what value of x you put there the function is expected to never go to the negative values okay of course a counterpart natural log behaves like this okay that's why they inverses so in this case what do we expect so meaning that what is determining the sign of the or any outcome of this function is purely what is in the brackets which is 1 minus X so we understand that if you get values of X that are less than one all values of X that are less than one all you're going to be getting is so if you have X being less than one values that you're going to get will remain positives and then if you get a value of x that is greater than one you now begin getting negative for example if you put a two there you'll get what you'll get a negative right yes so in a case where we getting a negative where X is greater than one we can here it indicate that within that domain the function is said to be decreasing and then in a case where the values of X are less than one we always get positive so the function is now increasing in such a case remember always from what is outside you always be getting a positive so what would determine the overall sign is what comes out from the brackets okay that's how you can look at such a function okay so of course I've mentioned we can also have cases where you have now a function which is I think we've already encountered this yeah I've already encountered this but of course let's try to look at this quadratic function let's try to differentiate it so y Prime will be 2x + 4 so for this function now this is why it becomes interesting because how do you basically know or tell when we expect this function to be increasing or decreasing so let's try by guessing first of all and then I'll show you the simpler way of undering it so by guessing when do we expect the the what when do we expect the Dy DX of Y Prime to be positive and when do we expect it be negative okay so clearly if I put a two there or if I put a negative two I expect this is going to give me zero right yes so it's like -2 is like the deciding point because that's why it's equal to zero so what if I try to get a value greater than -2 maybe a zero a one a two and so on and so forth so in such a case if I get for example a two I put it there I'll get a positive result so going this side I'm getting a positive result if I go below -2 for example -3 -3 * 2 gives me -6 + 4 which is a negative so going this side I'm getting a negative so I would guess to say okay well each time my value of x is less than -2 what am I getting I'm getting a function that is decreasing and then each time my value of x is greater than -2 what am I getting an increasing Behavior so that's by guess workor now the simpler way of trying to approach this is where we introducing what we now call the critical point so the critical point is where you expect that the dydx will be equal to zero so 2x + 4 becomes so equ whatever you deter as X equ to Zero so 2x = to 44 so X becomes -2 so that is always very important so of course we look at cubic functions and other bigger functions where you are going to have more than a single critical point and that is very necessary for you to determine the parts where the function is expected to increase or decrease so if you have a critical point like this one all you just have to do is try to now determine the behavior on the left and the positive side of which we've already determined that of course when you go to the side where it is less than -2 it's decreasing increasing on the other end now there will be interesting parts now where we have a cave like that so such a cave is going to have critical points at that point at that point that point so the way you get basically get to look at this is for example you would get this value of x there may4 and then one and then of course this is going to be done theoretically because it will not obviously be sketched so we'd have to look at4 and one and then you get any value plug in in the derivative function if it obviously you expect that here it to give you a positive value here it to give you a negative value okay it to continue negative value and then positive value so of course there are points where now at this points now these points where which we are going to talk about ACC stationary points where now even the YX will be equal to zero okay Point like that one okay so you understand but for now understand that will be using or determining what we are calling critical points to help us determine where parts of the function of a cave where it is increasing or decreasing and I think that is one of the co uh key Concepts that helps us to sketch a cave okay let's proceed now and just get to look at the stationary and the critical points let's talk about the stationary and critical point so it's very interesting actually I think uh I've never given it a thought not until I I was thinking about making this video about that so actually I've always known that when it comes to determining the critical points or stationary points it's of course a point where we expect that the Dy DX is equal to zero well while there's nothing wrong with that I didn't understand where we station came from so let's consider this graph maybe let me draw it more carefully so that you can see certain points that I want to be visible um I'll draw a first part of the graph like that be let me end it like okay then I'll begin in like this okay so I think that's that's what I want so when you look at this graph this part is increasing that part is decreasing this part is increasing this part is reducing okay so there's what we call increasing at reducing rate and all that you talk we talk about it as we move so this is increasing at a decreasing rate this part increasing at a decreasing rate because we are leing at a maximum so it's increasing at a decreasing rate and then this is increasing at an increasing rate okay so a basic idea is if I give you an examples of an exponential function which increases at an increasing rate and then also a natural log which increases at a decreasing rate it's um more like this so it's not as increasing as fast but it's increasing so increasing at a decreasing rate and then this this is increasing at an increasing rate okay just know that for now that there is also that so I've already mentioned to say this part you can see the graph is increasing or reducing and this part is still reducing okay so don't worry about that for now so if you look at this point we draw a tangent there and then we can draw a tangent there so so these points where the dyx is equal to Z or where the tangents to the cave are parall to the x axis if you want other terms but what you need to understand is at that point the YX is equal to zero so at that point the graph is said to be constant it is neither decreasing nor increasing so we call it a stationary Point stationary because the function is neither increasing or decreasing so it's like stationary stationary means no change okay so at that point so at these points where YX is equal to Z stationary points because the dydx is equal to zero or in other terms the function is never increasing nor decreasing okay now not only do we have that point we also have um the point where the the decreasing part the decreasing parts are meeting one from the minimum point the other one from the maximum point so where there's that change of behavior so more accurate what happens is this part is coming from the from the the maximum part like this and then another part going to the Minima comes to join so there is this part which is of course decreasing at an decreasing rate and then there's this part decreasing at U approaching the minimum so at this point where that there is that change you can see there's more like a cave okay like this um like this and then that's what done anyway this is what I'm trying to talk about the point where that occurs is called a point of inferion so even at that point the YX is equal to Z okay so a point of inferion so a function is never maximum or minimum at that point but YX so when YX is equal to Z and then the point is neither a Maxima nor a Minima point the point is called a point of inferion so of course we look at the second derivative the second derivative is what helps us to determine whether a point is a Maxima or a Minima so of course when the second derivative yeah when the second derivative is greater than zero or in other terms positive the point is said to be a Minima when the second derivative is less than zero meaning that it's negative it is a Maxima but of course in a case where it is equal to zero that's now what we call point of inferion so stationary points are points where the Dy DX of the derivative is equal to zero and we've got three examples of stationary points maximum Minima and the point of inflation okay now what are critical points so what you need to understand is the stary points we've talked about are critical points okay so critical point is these are points where the dyx is equal to Z or the function does not exist or equivalent we can say or the derivative does not exist at that point so meaning that critical points is a bigger set consisting of stationary points as well I can give you an example of such a cur so we can take this the some some normal kind of a cave okay so of course I believe you've talked about certain points that may be undefined okay but of course being more realistic which usually happens in most of the cases where you approach uh zero at the point where X is equal to zero usually undefined but can be at any point okay so that would be in a case where on the denominator for example you just have X and then you have a certain function on top something like that so a case why X is is like equal to zero at that point you expect that the function is not defined this may be a derivative or just a function itself so at that point that point is is actually a point we can refer to as it can still be called a critical point so this may even be at the Minima Maxima it will be a critical point so what makes it different is that stationary points are basically seary points are restricted to a part where YX is equal to Z in a case where the derivative of a function is undefined those are still referred to as critical points so critical points is a bigger set broader as compared to the stationary points and I think this is the key difference that most students get to miss now let's look at a few practice questions that will just help us solidify whatever we've talked about so consider a case where f of x is = x^ 3 - 3 x 2 - 9 x + 5 so find the all the critical points of that function how would you do it so of course we expect to differentiate the function so we're going to have 3x^2 - 6 x - 9 so that is our derivative function so for us how do you find the critical points so we said the critical points of course whether they still critical points or stationary points this you expect that the intersection part is where the dydx is expected to equal to zero so equ the derivative function to a zero and then try to find the value of x now you can clearly see that this is now what that is now a quadratic function that you can easily look at so three is common you can factorize it to make it simpler so what we can remain with what x^2 - 2x - 3 = Z and then we can factorize what is inside of course so if we do factorize what is inside there we're going to have x +1 x - 3 = to Z which can easily help us find the values of X so x isal 1 and then X is equal to positive3 you can solve that using any formula you want either by factorization as I've done it by formula or by completing the square so the critical points can be found by you can find you have to find the Y values so only you found the X values so 1 comma something 3 comma something so what you're going to put there is now for you to get the Y value you need to go back to the original function plug in the value of what plug in the value of the x that you found so for negative one if you pluging 1 1^ 3 is1 and then that is - 3 POS 9 + 5 9 + 5 is 14 14 minus 4 I believe that gives us a 10 and then all the same if you plug in a three there you get -22 so that's how you find the stationary or the critical points given a simple curve like that one let's take it a bit more interesting and then get to look at something more advanced like that you look at an exponential function so of course we've said for you to handle any of these you need to first of all look at what you need to look at the AO finding the derivative function so F of inverse of 2 is going to be how do you differentiate an exponential function so multiply by derivative of the power natural log of a base so in this case the derivative of3 T is3 and then multiply the origin of function itself and then natural log of a base which is just natural log of v in this case so that means that will disappear I'm trying to mention that because you may have a case where of 2x how do you differentiate that derivative is Der of the power which is a one maybe if I put is to^ 3x the D of power is 3 m by the original part and then natural log of two now for e natural log of e is just one so that's what we have and then derivative of 2 T is just two so at that point we've differentiated the function now for us to determine the for us to determine the critical points we need to equate this to zero right + 2 is equal to Z so how do you basically get to such a point where this is going to be equal to zero so obviously we can take the constant other side e 3 t plus becomes -2 the other side and then of course if you divide by the negative of three both sides e what are we going to have we're going to have 3 e or e 3 is equal to 2 over 3 so how do we get to find the value of uh of T what's the best way for us to find the value of of that so the the way would handle that is by introducing the natural log on both sides since we know that it is basically the inverse of e so we are going to have3 T isal to natur of 2 3 so natural log of 23 is something that you can actually simplify using a calculator so of course you can divide Again by -3 both sides so you're going to have1 3 and then natural log of 32 so you can simplify that on the calculator so what you're going to find there is a value of T which is more like the value of x and then to find the value of y you would have to go back the function and substitute there yeah that's how we going to find the critical point I'm I just needed to mention that that of course it's not as complicated as uh as as it sounds okay so no matter what you're given if you want to find the critical points all you just have to do is equate the derivative function to zero so let's consider now one more interesting form of the function that I want us to to look at so what if f ofx um is equal to X and then ra the power two over the three feel free to just pause the video and just St determine the critical points the derivative becomes 2 over 3 and then X reduce the power by one so 2 over 3 minus a one that's going to be what negative 1 over 3 now some of you know that if you've got X ra the negative power you can just take it the bottom so that you have 2 over 3x and then ra the positive power of one / three okay so how do you find the critical points of this function so of course we've already established to say you get to find the critical point where the derivative is equal to Z now you notice one thing about this is if you have got two over 3x^ 1 3 this can never be equal to zero no matter what you're going to do it can never be equal to Z meaning that in this case you don't have what stationary points no stationary points but of course you know that if you plug in a value of zero there X becomes zero this thing becomes undefined right so now where the functions are undefined that is referred to as what a critical point okay so a critical point is where the derivative of a function is undefined in this case this is the derivative so the derivative is undefined when x equal to Z so it's the critical point becomes Z comma you find the y coordinate by going back to the original function which is 0 set Power which is still Z so that is a critical point so this is what differentiates now the critical points from the stationary points clearly with such an example so for the sake of your practice I want you to just try out these two two interesting questions consider a case where Y is equal to X e3x determine the critical points and then one more would be x + 1 /x determine the critical points