this video has differential calculus topics and remember you can subscribe to the teach quant youtube channel and the number of topics that you're going to consider in this video the first one is the properties of limits the second one is how to find the limit of a function followed by how to differentiate differentiation by product rule differentiation by quotient rule how to differentiate using chain rule how to differentiate a parametric equation application of differentiation how to find stationary points and we compare how differentiation and integration differs all how they are related so these are the 10 different videos that have been combined together so that we can have one video for what you'd call differential calculus so i hope you'll enjoy the videos properties of limits and we want to look at some examples assume that a limit as x tends to a of f of x is equals to 4 and limit as x tends to a of g of x is equals to negative three for example if we have been given limit as x tends to a of f of x plus g of x co according to the properties of limits this is limited as x tends to a of f of x plus limit as x tends to a of g of x and we know that limit as x tends to a of f of x we've been given as four and this one has been given as negative three and that gives us the answer to be equals to one another example limit as x tends to a this is now where we have a quotient this we can write it as limit as x tends to a of the square root of f of x divided by limits as x tends to a of g of x and that basically takes us to the square root of 4 all over negative 3 which is negative 2 over 3 limits has x tends to a of 2 g of x will be equals to 2 limit as x tends to a of g of x which is equals to 2 times negative 3 which is equals to negative 6. the fourth example limit as x tends to a the cube root of f of x minus g of x is equals to limit as x tends to a f of x minus g of x everything raised to power 1 over 3 which will basically be equals to limit x tends to a of f of x minus limit as x tends to a of g of x everything raised to power negative and we know that we have been given the limit of f of x and g of x so this is 4 minus minus 3 raised to power at that which is 7 raised to power at that to find limit as x tends to 2 of 2 x squared minus 5 x plus four this will be a direct substitution so it'll be two two squared minus five times two plus four which will be 8 minus 10 plus 4 which is to be 12 minus 10 and the answer will be equals to 2. limit of a function and basically we want to look at a number of examples and one of them is limit as x tends to 2 of 10 and that basically remains as 10. what about limit at x tends to 2 of x it will give us 2 because what are we doing the idea behind what we are doing is uh if we look at limit as x tends to a constant c of f of x then it should be f of c so we replace uh the f of x x with c what you have done in that uh example and that's how we would get the limit of a function and that means if we have another example and in this case i want to get a limit as x tends to three of x squared plus one basically this will be three squared plus one and that will give us nine plus one is ten what about in cases like limit x tends to 4 of 4 over 4 x squared that will be four over four squared and that will give us one over four example number five if i was to get the limit x tends to 1 of x minus 1 over x squared plus x minus 2. now you note that if we had to replace this dialect it will be giving me 1 minus 1 that 2 upon squared plus 1 minus 2 which will be 0 over 0 and this is what we call as the in d ta minutes form and so what when we have in the terminate form that's when we use what we call the el hopitos law and so what do we do we differentiate the denominator and the numerator at the same time and then we substitute so i'm going to go through the problem again so limit x tends to one so when i differentiate the topmost part it will give me one and when i differentiate the lower part it will give me 2x plus one and when i replace one it will be one over two plus one which is at that so that's how i would approach that when i get there in that indeterminate form of the problem that i have and that will apply in other cases for example uh for example a problem like the following we want to look at we look for limit as x raised to x x tends to zero if i was to substitute this directory it will be sine 0 over 0 which is 0 over 0 which is again indeterminate and since it is indeterminate and then what do we do so it is in the terminate so what we do is we get limit as x tends to zero over cos x over one and if you substitute it will become zero over one is equals to one so whenever you have the indeterminate form then what we use is what we call the l cos optos rule there are so many other examples we can think about we can think about this where for example i have uh let me say it's infinity so if i substitute this again it will be infinity over infinity and the infinity over infinity is uh is not known so what we do is we go through the l orbitals load again and then it basically means then then we need to differentiate to x squared which gives us 2x and e x again remains as ex if i substitute i'll get 2 times infinity over e list to power infinity and again that means we repeat the process because it is giving us indeterminate form i also give you a problem so it will be 2 over e list to power x and that will be true a raise to power infinity which is 2 raised to power 2 over infinity and any number remember that uh uh that one over infinity is equals to zero so if you keep that in mind then that should be the answer so you note that you have differentiated this twice for us to get the limit what about if you have other problems like for example this if i said let's get limit x list to x to five x over x minus five now even if we had to differentiate uh even if we had to substitute we'll get five over zero uh which is undefined so remember this is not indeterminate because it is not zero over zero it is undetermined and undefined so what we say is that limit does not exist so that is something that we want to keep in mind and uh we can go on and on and on using other approaches we can think of other problems uh i can give you maybe two more so if for example to get limit as x tends to infinity of x squared plus x over x squared definitely if you look at that problem uh you realize that if i was to substitute directly it will be infinity over infinity so what i'll do is i'll be slightly clever and i'll rewrite it in a different form i'll write it as 1 plus 1 over x and that takes me to 1 plus 1 over infinity and that tells me that the limit for that would be equals to one i could have still use the l options law because infinity over infinity is also tricky if i was to use the l orbital's law then it will be limit as x tends to infinity i differentiate the power at the the the the numerator will be 2x plus one i differentiate the nominator it will be 2x then again i go again and differentiate the second time it will be 2 over 2 and definitely that one will be x tends to infinity of a constant just like our first problem and that means then i'll stop at that point so these are some of the examples of how to use limits to consider a few examples perform different types of differentiations so if we are given an example of y is equals x cubed so the first derivative of y with respect to x which we can write as d y over d x all you can write as y with a prime what do we normally do you take the power down which is three and then you decrease the power by one so the answer gives you three x squared i want to get second derivative which i can also put down as y double prime or if you wish you can write it as d y squared over d dx squared so i'm going to differentiate the the solution i got in the first step so i put 2 down and i decrease the power of 2 by 1 and that gives me the answer as six x again the y prime i'll write it as y prime which will be four then i'll take four and then i take the power of x minus one so that will give me 16 x less to power 3 and that would be the solution to then we differentiate that now with that idea i take 2 down and i reduce the power of two minus one gives me one i put two on the side remember the power of x is one so it will be one minus one and they are literally what it means is i'm multiplying with x raised to power 0 because what we know that any number to power 0 is always 1. and if you allow me then i would put 1 down i'll put x and then i'll subtract by one so what am i trying to say if i have a constant and i try to differentiate the constant the answer is always zero this will be 2x x raised to the power 0 we have seen any number list to power 0 is 1. so in the second part here x raised to the power 0 is 1 so it will be plus 2. and then here it will be 0 times 1 times x raised to the power -1 and everything remains 0. so the first derivative of that polynomial would give us that what we do is we take this and we differentiate we take the power and then we differentiate so it means in the first derivative well because when you differentiate x you get one you put the one there and then you multiply with the original function as it is so that the first derivative of a less power x remains e raised to power x again you take the power you differentiate and this gives me two x plus one so after i differentiate the power so what is the solution the first derivative will be equals to 2 x plus 1 and then i write back the original function as it was then we take this we differentiate it so when we differentiate then the solution becomes after differentiating x you get 1 and then you divide by original variable which is x so the first derivative of ln of x is 1 over x so the first derivative what do we do we take 2x plus 1 again this one when you differentiate 2x plus 1 you get 2 after you get 2 then you divide by original function so once you differentiate you put the value there after you differentiate after that after you differentiate you put the value there then you divide with the original uh variable as it was so that and y prime is equals to cos x that is uh and then y is equals to cos x and y prime is equals to negative sign x so we consider the product method of our differentiation and i want to get the first derivative of that so what does the product's slow says it says that uh i like itemize my variables in terms of u and v so what do i do i differentiate you i keep v constant and then i keep you constant and then i differentiate b so in this case i differentiate ln of x which will give me 1 over x then i keep e x constant as it is in the second case i keep ln of x constant and then i differentiate e of x and when you differentiate e of x we still have e of x so that's the solution for my first derivative of that so i could write it as that plus that and that's the first derivative when i use the product law another example i could think about when i'm using a product rule is uh an example of the form y is equals to x raised to power 4 maybe e less power to x again i separate it into two as u and v and therefore y prime is u prime v plus u v prime meaning u prime means the first derivative of u with this vector so i'm referring to that as my u so my y prime will be four x cubed e of two x plus x four times 2 e of 2 x and therefore i can write this as 4 x cubed plus 2 x list to power 4 and i could leave that as a solution to that product and therefore the first derivative is i differentiate sign i get cos then i multiply with cos x and then i keep the sign x and then i differentiate because i get negative sign x so that would give me that and uh you go ahead and uh multiplying them out therefore that will give me cos squared x minus sine squared x so what if i want to differentiate this the first part here i have two parts product rule so the derivative of this will be two x times five x squared plus one list to power four ln of x plus x squared plus 1 raised to the power 5 times 1 over x and that gives me uh the derivative now what what have i done uh the derivative of ln of x is one over x the derivative of x squared plus one plus the power of five is uh you take five down you reduce the power by one that gives me x squared plus 1 to the power 4 and then you differentiate what is inside the bracket x squared plus 1 when you differentiate you get 2x so the first derivative of that therefore i can write it as x squared plus 1 to the power 4. and then 10 x lean x plus 1 over x x squared plus 1 yeah and you can differentiation by quotient rule and the general formula that we normally use is given by y is equals to u over v so we assume you've been given an equation of that form and therefore the first derivative the general formula that we use so here we develop a formula is we differentiate you you keep v constant minus u times the first derivative of v and then we square the v so we know we want to make one our u the upper term to be our u and the lower term to be our v so mostly the quotient rule means we are dividing one function with another so if my u is equals to sine x the first derivative of u with respect to x will give me cos x and if my v is cos x then my v prime will be negative sine x so according to that formula then y the first derivative of y is u prime which is cos x times v and that gives me cos squared x then minus u and v prime u and v prime is another minus sine squared x and that and then we know what is v we know v is cos so yeah cos then you square it is cos squared so you see it's cos squared minus minus which is y positive so cos squared plus sine squared is one that is a trigonometric identity over that and therefore the first derivative of tan x is equals to sec squared x all is it now the trick i would have used here is i would have gone ahead and wrote it down like this and from the knowledge that i have i would have done this and i would have gotten the answer as that which is the same as minus 2 over x cubed i could have done that from what i know but i could also as well have used the quotient rule method and uh so my u is one my u prime is zero my v is x squared and my v prime is 2 x so my y prime would be u prime 0 times x squared minus 1 times 2 x divided by x squared which is minus 2 x over x squared squared which is x to the power 4 and you can see already the solutions are equivalent is minus 2 over x cubed so even if i use the quotient all i could have done it directly i would have come up with the same solution will be one my u prime will be zero my v will be cos x and my v prime will be equals to negative sign x so that's what i would get so what is the derivative of set x the first the first derivative of sec x will be u it will be 0 times cos x minus u times v and then you divide by cos squared x and so that's what you get and that's that's zero that will be sine x all over cos squared x so the first derivative of sec x is sine x over cos squared x or if you wish it could have been sec of x times tan of x because how did i know that remember this is the same as sine x over cos x times one over cos x and that will give me one over cross is sec and then i have the turn u is ln of x and u prime is one over x so that's the first thing you have and then we have v is equals to x squared and v prime is two x so the first derivative of y will be u times u prime times v then u times v prime 2x times ln of x and then everything we divide by x squared which is our v and then we square everything so that's what we have which will be x minus 2 x ln of x all over x to the power 4 which will be 1 minus 2 ln of x all over x cubed and that transition by chain rule is the method we want to consider so what we do is we let u be equals to x squared plus 2x plus one and therefore my equation changes to y raised to power six and uh now looking at that equation i would not be able to look for the y dx i'll literally be looking for divided u so for me to be able to look for d y d x which is the first derivative of y with respect to x i'll try to find the d y d u and then i look for the d u the dx now if i differentiate d y with respect to u i'll get six u raised to power five and then if i differentiate u with respect to x i get 2 x plus 2 and so that is what we call the chin it makes it simply for simpler for you to differentiate substitutions and therefore my final solution becomes 6 2x plus 2 and then we put back the u that we add and remember u is x squared plus 2x plus five uh u is x squared plus 2x plus one list to power five and that gives me my the first derivative of y with respect to x i can then i let u is equals to 2 x plus 1 therefore my equation becomes y is equals to sine u why do we do this to simplify the part of sine now it's only u so if i going to try and differentiate so i'll get the y with respect to u and then you with respect to x that's so divided you the derivative of sine x is cos so it will be the cause of u for the first part and the second part will be two so the first derivative which is uh the y dx the y prime will be equals to two cos over 2 x plus 1 and so so what i do is i let u be equals to cos x therefore y becomes e list to power u and again therefore d y d x d y d x is the y d u because now y is related to u times then d u over du dx because u is related to x so and that's the method we call the chain rule method so if i differentiate the cosine uh what do i get first of all d y d u is e less power u and then the d u d x is negative sine x so the first derivative will be negative sine x e l is power u and remember our u is cos x now the same problem we could have solved it uh we could have decided to differentiate it without employing the use of the chain and for those of us we could have taken this you differentiate it and that will give us negative sign x and therefore you could have said that the first derivative will be equals to negative sign x then e less the power the original function and you can see those two solutions are exactly the same parametric equation is what we want to look at and then y is equals to cos 2 t and x is equals to sine t then we are told to find d y over dx now you'll see that y is represented in terms of t and x is represented in terms of t but then we're supposed to find d y r d x and so d y d x i can see it is the same as d y over d t because we don't have a direct differentiation of the y dx and then dt all over dx that's how we are able to link the two together so that we are able to differentiate now what is the first derivative of y with respect to t this is uh we see y is equals to cos 2t so what i'm going to do is i'll say y is equals to cos 2 t if i let u is equals to 2 t then y becomes cos u now what am i interested in i'm interested in finding the y d t so d y d t is equals to d y over d u times d u d t and we know that d y d u is a derivative of cos which is negative sine u times d u d t is 2 so that gives us negative 2 sine u so the first part of d y d t is negative 2 sine u and you know what is u u is 2t so because negative 2 sine 2t what about dt dx remember what you have been given is x is equal to sine t so the first thing is i love to look for the x dt and the first derivative of x with respect to t is cos t and therefore dt over dx is 1 over cos t from that i just get the reciprocal reciprocal of of that value that we got so d y d x will be equals to d y d t times d t over d x which is negative to sine 2t times 1 over cos t so that would give me the solution to that faster derivative of 2 sine 2t divided by the cos of t so that is a solution that you could leave it there for some of us who are familiar with the double angles we could have taken that problem to another level and the level we are talking about is sine 2t can also be written as 2 sine t cos t and so if i could write that based on the principle of the double angles or the angular identity then d y d x can be written as negative two times 2 sine t cos t all over cos of t which will give me negative 4 sine t and that will give me the final solution we have been given x is equals to 3t and y is equals to 90 squared so the first part is to find the y over dx again the procedure is the same and what does that mean for d y over dx is equals to d y d t times d t over d x d y d t is 18 t 18 t and dx over dt is equals to 3 and therefore because we are we need ddx not dx dt therefore d y d x will be equals to 18 t times one over three which will give me the solution of 60 for that case because we need dt dx which is the reciprocal of what we found there find d squared y all over dx squared now what does that mean it basically means d squared y all over d x squared is to find d dx of d y d x and that is fine the d the x of 60. now what is easier i could use the implicit method or i could say that since x is equals to three t then two x will be equals to sixty i just multiply by two so whenever then at least a 60 then i'm going to replace it with 2x that's basically what i've said there so then i could say d dx instead of 60 i'll say 2x and that should give me a solution of 2. that would be the answer for that now if i wanted to use the implicit method i could have said alternatively d dx of 60 will be d dt of 60 times dt all over dx and you remember we got dt all over dx we know that if x is equal to 3t then the x dt is at that and therefore i know that this will be ddt of 66 then times at that and it would have still given me the same solution just like the similar one that i have been able to get the implications of differentiation we have an example for a dosage of x cubic centimeters of a certain drug the resulting blood pressure is approximated by b x is equal to three or five x squared minus eighteen thirty x cubed and x lies between zero and 0.16 so we are trying to find the maximum blood pressure and the dosage at which it occurs so differentiation so we get the first derivative which gives us a 6 10 x minus if we multiply that that becomes 54 90 x squared the b prime of x is equals to zero and uh we want to get uh the critical point and so if we do that basically means that uh uh 5490 x squared is equals to 10 6 10 x and therefore x is equals to 6 10 x over 54 90 x and that gives us the value of x at 549. so this is our critical point so the critical point of us so the critical point is at x is equals to 61 549 which is basically the dosage that we are talking about yeah so we are told to find the maximum blood pressure and the dosage which it occurs so the dosage is uh that is the dosage because x is the dosage now let's find out if actually this is the maximum dosage that we can get how do we get the maximum dosage we test so we are going to differentiate our b two times we have differentiated the first time and this time it gives us 16 minus 10 980 x and when we substitute the value of x 61 over 549 it becomes 16 minus 10 980 61 549 and this value will be equal to 610 minus uh 549 goes to 1098 two times so it will be 20 times 61 and definitely is less than zero so so then then we know that bx as a local maxima at that value that you've been given x is equal 61 549 so to answer the question the first part is the maximum bp that is maximum blood pressure that will be getting from the computations it basically be taking our bx our bx our x is 61 549 and then we substitute it into this equation then minus 18 30 61 549 cubed and so the maximum blood pressure will be 1.255 and then the second part is the dosage the dosage is x is equal to 61 549 [Music] moving is given by the equation s is equal to 160 minus 16 t squared s is in meters and three in seconds find the body's velocity and the relation as time t we know that velocity is given by change in speed with time so that gives us the velocity the change its speed with time and that is what we have and the velocity is given by meters per second and then the second part is the saturation and acceleration is given by change in velocity with time and that gives us the acceleration and it is in meters per second squared and it's a negative sign because the object is decelerating coming i think equations of a curve and x is equals to three t over one plus t and y is equals to t squared over one plus t want to find the equations of the tangent and normal at the point for which t is equals to two which is a d y over d x or if you want to put it in a slightly different a it is changing y all over change in x and we know that d y d x will be given by d y over d t times d t divided by the d x over dt not dx the dy over dt what then we need to find out is uh based on our equation uh we are going to use the quotient rule our u is t our u prime is two t and then our v is one plus t and then our v prime is one so d y over d t is u prime v which is two t one plus t minus v prime divided by one plus t squared and x over dt uh remember that our u is given by 3t so our u prime is 3 we want to use the quotient rule to be able to differentiate this is one plus t and my v prime is one so my dx dt is u prime v minus v prime mu i think that's what we have divide everything by one plus t squared and this this simplifies to 3 plus 1 plus t squared now we are looking for dt dx dt dx will then be equal to 1 plus t squared over 3 and that takes us back to our problem that which is finding the slope d y d x is equals to d y d t times d t over d x and the first question is what is my d y d t my dydt will be 2t plus t squared over 1 plus t squared then times 1 plus t squared over 3 and that simplifies the equation to a form that gives me my slope of the line as 2t plus t squared divided by 3 and since i was thought to find the value at t is equals to 2 this will be 2 times 2 plus t 2 squared divided by 3 which is 8 over 3 and that gives me the slope of the line of the tangent first of i need to find the value of x when t is equals to 2 and that takes me to the equation of x which is 3 times 2 all over 1 plus 2 which gives me the answer is 2 and then i look for the value of y at t is equals to 2 and y is given by 2 squared over 1 plus 2 which is 4 over 3. now y minus 4 over 3 divided by x minus 2 is equals to 8 over 3. so 3y 3y minus 4 is equals to 8x minus 16. so 3y minus 8x is equals to minus 12 would form the target equation to find the normal and we know that uh for us to get the normal we know that uh negative three over eight times eight over three is equals to negative one so this forms the gradient when we are forming the normal because for that to be then at the normal then the product of the true should be goes to negative one and therefore using the same points y minus four over three x minus 2 should be cos to minus 3 over 8 and therefore the equation to the normal it will be 8 y minus that 2 over 3 is equals to minus 3x plus 6 so 24y minus 32 is equals negative 9x plus 18 and so that forms the other equation which is uh the normal equation 24y plus 9x is equals to 50. the normal equation that we get we look at the relationship between differentiation and integration and basically what happens is that if we are given a function here and this function what happens is when you differentiate what happens is that you get the first derivative if you have been given the first derivative and you integrate it then you get the original function and this is basically part of the key lesson that we learn in calculus when you're f of x to be equals to x squared therefore the first derivative of that will give us 2x so for us to get the original function f of x will then be equals to the integral of f prime of x dx and that is the integral of two x dx of integrations will get to x squared over two plus c which is x squared plus a constant we want a relationship number two we want to look at the relationship between tan x times sec x and sec x so if we integrate tan x times sec x we get sec x and then we differentiate sec x we get tan x factor x square consists then df dx will be equals to what and we expect the answer to be equals to tan x sec x remember your f of x if i'm differentiating i can write this as 1 over cos x and this i'm going to use the quotient law so 1 my u prime is equals to 0 my v is equals to cos x and my v prime is equals to negative sine x and therefore my df dx df over dx is equals u prime times v minus minus sine x divided by cos of x squared using the quotient rule which is sine x all over cos squared x and that gives me tan x sec x so the integral of tan x sec x dx will give me sec of x plus c just the constant of integration so how do i do that i'll take this and i'll write it exactly as sine x over cos x times 1 over cos x dx and that will be the integral of sine x all over cos squared x dx so i let u is equals to cos x and i want to use a substitution method then i say d u my d x is equals to negative sine x and that means that my dx is equals to minus d u all over sine x and so that means then i can write the equation as sine x over u squared times negative d u all over sine x so what i'm sure about is that these two will cancel out and uh that makes it easier for me with that information then what i have is a negative u raised to power minus 2 plus 1 divided by minus 2 plus 1 plus c so and that gives me 1 over u plus c and you know what is 1 over u 1 over u is 1 over cos x plus c which will give me the sec x plus c so i've shown that divide if i have a function f of x is equals to 2 x all over x squared plus 1. so what happens to this variable it's it's it is 2x over x squared plus 1 when i integrate i get and when i differentiate so this is when you differentiate and this is when you interpolate [Music] so that's what we want to show that when i differentiate i get that and when i integrate what do i get so that we can link the relationship so start with the differentiation f of x is ln of x squared plus one so the df over dx will be equals to so how do we differentiate the ln you take this you differentiate it and this gives you 2x so i'll put 2x there and then i'm going to divide with the original function and that is the rule of differentiation so i've seen that if i differentiate line of x squared plus 1 i get 2x over x squared plus 1. what about if i do the integral part if i integrate 2x over x squared plus 1 dx so i let u is equals to x squared plus 1 and then dudx is equals to 2x and what does that give you and that gives you tells you that my dx is du over 2x so if i come back to my integral i'll write it as 2x over u times du all over 2x again we can easily see that that cancels and if i integrate u over u the integral and this gives me ln of u plus c and that is ln of u my u is x squared plus 1 plus c and that's your want to look at the stationary points and in the figure there we can see that we have a point a which we call them local maximum we have the point b which is a local minimum and the point c which is our shadow point and the three points point a b and c are what we refer to as the stationary points so given the function y is equals to x cubed minus three x plus three find the stationary points so we want to use the principles of differentiation to be able to do that so what find first derivative and that basically means that d y d x is equals to 3 x squared minus 3 and that we are going to equate to 0 to be able to solve for that and this gives me the equation of the form 3 x minus one x plus one is equals to zero means that x is equals to one all x is equals to negative one when x is equals to one y is equals to one cubed minus three times one plus three and uh that gives us 1 so when x is 1 then y is equals to 1 and when x is equal to negative 1 y is equal to negative 1 cubed minus 3 times negative 1 plus 3 negative 1 cubed is negative negative 3 times negative 1 is 9 3 plus 3 is 9 so y is equals to negative one plus three plus three which is equals to eight so we have our stationary point one of them is when x is negative one y is eight and the other one is when x is one then y is one so didn't actually call them uh stationary points i'll i'll be calling them uh turning points because we don't know are they maxima are they minima or are they sandal point so we use the second derivative as the point so the first the first is x is equal to negative 1. divide the disc at y over dx squared is equal to negative 6 and negative 6 is less than 0 then that tells us that that point is a maximum point if the second derivative in the second case is equals to six which is greater than zero then this is what we call the minimum point in the case that the second derivative would have been equals to zero then that would have been a certain point so i have two points i have a minimum and a maximum point so those are the two points i have one one is a minima and negative one eight is a maxima so with that information now i can plot the class now based on the information that we have i've been able to make a sketch whereby i have the two main points i have my maximum point here and i have my minimal point here and where the line cuts through the y axis and i've been able to make a sketch using that set of information so that is the process that we normally take when we want to get the minima the maxima and the saddle points on what we call the stationary points