in section 3.3 we begin our study of the rules for differentiation so now that we have a conceptual foundation for the derivative this is a process that we would like to speed up so you've seen that we can find the derivative by definition but it's not necessarily easy and it's not necessarily fast so we've seen some cases where the derivative by definition is doable and we have that foundation now for this understanding of what this function called the derivative is but going forward we would like to be able to speed that process up and work a little bit more efficiently so in this section and pretty much the rest of chapter three we're going to require acquiring a set of derivative rules so part of advanced mathematics is not necessarily working harder all the time sometimes we like to recognize and exploit patterns and that's what we're going to be doing through the rest of chapter 3 here so we're not in we're not going to be finding the derivative by definition any longer so that is limited to sections 3.1 and 3.2 we want to remember the definition we want to remember what's under the hood of all this this process that we're doing of differentiation and finding the derivative but we're going to move away from working through it explicitly through its definition and so in this section we're going to begin to acquire some derivative rules and we'll be acquiring more and more rules as we work through chapter three so first off here the derivative of a constant function is always going to be zero so a constant function is a function perhaps of this form f of x just equals some number so here f of x equals c where c is a real number that derivative is always going to be zero as we can see by applying the definition we end up with the limit as h goes to zero of zero which is of course zero so we can think of the derivative being as the instantaneous rate of change well a constant function never changes so we would certainly like to see that the derivative is zero and that is the case so the derivative of a constant function is 0. next we have a linear function so a function of the form mx plus b we can go through the definition we find that the derivative is equal to m and m is the slope right so that makes sense the derivative of a linear function its instantaneous rate of change well the instantaneous rate of change is going to be the slope in this case so now moving forward to what we might call power functions or power terms so a function with a variable raised to a power so for instance here x to the n n is just any real number so what we find is that the derivative of this power term x to the n is the expression nx to the n minus 1 power so this is called the power rule so what happens is this power of n comes down to the front it becomes multiplied by the variable x and then we subtract one away from the power so the derivative of x to the n is going to be n x to the n minus one so this is often referred to as the power rule so as a note here think of the function y equals x to the one thousand power well by definition we're stuck with this limit and this limit would be extremely difficult to work by hand uh because of this binomial here being raised to the power of one thousand so pretty much impossible to work this def this derivative by hand but by the power rule we've got the result immediately the derivative of x to the one thousand is going to be one thousand times x to the 990 power so the 1000 comes down we take one away and that becomes our derivative so quite amazingly these power terms the derivative has a very beautiful pattern that comes out in finding their derivative is just bring the power down take one away and that's the derivative of a power term through what we call the power rule all right if you're interested here's a proof for this for the power being a natural number not something you're expected to know just for your curiosity if this is something that interests you so you can pause here and try to follow the argument for how we get the derivative of x to the n being an x to the n minus 1. so we've talked about the property of linearity as it pertains to limits so naturally since the derivative is a limit we expect the derivative to be linear so what we do find is that differentiation is a linear operator so in other words the derivative of functions being added or subtracted and then multiplied by particular uh constant coefficients what we have is that we can separate the differentiation process term by term and the constants just come along so the derivative here of s times f of x plus or minus t times g of x is just going to be s times f prime of x plus or minus t times g prime of x for any s and t being real numbers and this is just a consequence of the definition so we don't really necessarily think about this a whole lot other than the fact that constants come along with each term we take the derivative on and what this means is if i have a function that has many terms being added or subtracted i can just take the derivative term by term and then the constant any constants attached to those terms come along so let's use some of these rules that we have now to find some derivatives so first is the function f of x equals 2x squared plus 3x so the derivative very simply is going to be f prime of x so term by term i have 2x squared so what's going to happen is this 2 is going to come down i'm going to have 2 times 2 which is 4. we're going to take 1 away from the power so 2 minus 1 is 1. so this term has a power of 1. okay i have now plus 3x now this is a linear term so the derivative of a linear term is just going to be the coefficient so this is just going to be plus 3. and that's also the power rule so what i have here is a power of 1 the 1 comes down and gives me 1 times 3 now the power of 1 subtracts 1 so i get x to the 0 which is just 1. so i could equivalently write this expression 3x to the zero but x to the zero is just one so we typically will not write that so the next function i have here is x to the fifth plus x to the fourth plus x to the third plus x squared plus x plus 1 so term by term the derivative of x to the fifth is going to be bring the power of 5 down take 1 away so 5x to the fourth the second is going to bring bring the power of 4 down take one away four x cubed the third x cubed will bring three down take one away so i get three x squared the derivative of x squared is going to bring the two down take one away so 2x to the one power and now the derivative of x is just going to be one so that's a coefficient of one attached you can think of this as being the linear case or the power rule one comes down take one away i get the power of zero and then finally the last term is the constant its derivative is going to be zero so we typically won't write the plus zero at the end but you can if you'd like to and the last function here i have four x to the fourth power plus 3x cubed plus 2x squared plus x minus 1. so term by term the 4 comes down i get 16x cubed the next term i'm going to get 9x squared the third term i'm going to get 2 comes down i'll get 2 times 2 is 4 take 1 away so i get the power of 1 and then plus x gives me the derivative term plus 1. and then finally the negative one at the end its derivative is zero since we have just a constant in that case so notice how much quicker this goes now you should remember that at the end of the day here this what we're doing is we're effectively just jumping to the end result of that limit so again this derivative is the limit as h goes to zero of f of x plus h minus f of x divided by h but now we know the end result without having to do all that work in between so the end result of that limit in this case this first exam first example would be 4x plus 3 and the second that end result through a lot of work would be this result here and the same thing here in the last example so again remember what's going on underneath everything effectively now we have a very quick way of evaluating that limit this represents a limit so we don't want to forget that this represents a limit although we're moving away from expressing it in that way so so keep in mind what is going on under the underneath all of this uh effectively again we're still just finding a limit now we're becoming fast at it so the power rule also applies to negative and fractional powers so in particular remember that your roots are represented by the fractional powers so for instance the b root of x over a is going to be x raised to the a over b power so this is still the power rule the power of a over b comes down and we subtract one away and we can simplify so last last time we showed by hand the derivative of y equals root x is y prime equals 1 over 2 root x so by the power rule we get the same result so root x is x to the one-half applying the power rule the one-half comes down we take one away so that's one half x to the negative one half which now simplifies to one over two root x so the power rule applies to fractional powers as well as negative powers so a variable term to any power positive negative fraction is going to be addressed by the power rule and you could even say powers of 0 as well but in those cases powers of 0 are just the value of one so let's find the derivative of a couple of these functions that are going to have negative and fractional powers so here i have 1 over the function y equals 1 over x minus 1 over x squared plus 2 over x cubed minus three over x four x to the fourth so we can rewrite this as x to the negative one minus x to the negative two and then this is going to be plus two x to the negative three and this will be minus three x to the negative four so the derivative very simply now is addressed by the power rule so the first term we bring the negative down take one away the second term we bring the negative down that's going to be giving us now a plus two so we'll have negative there's a effectively a negative one coefficient here so negative two times negative one take one away the next term will bring the power of negative 3 down that's going to give us negative 6 x to the negative 4th and then finally bringing down the negative 4 power we'll have plus 12 x to the negative five so notice that these coefficients come along effectively i have a one here i have a negative one here a coefficient of two coefficient of negative three those coefficients come along with this process of differentiation and so in these cases these powers are being multiplied to the coefficient and this is how i'm getting in this case this coefficient and the derivative of plus 12 this coefficient of negative 6 this coefficient of positive 2 and this coefficient of negative 1. and if we wanted to we could go ahead and convert this back into we could eliminate the negative exponents this would just be negative 1 over x squared this will be plus two over x cubed this will be minus six over x to the fourth and this will be plus twelve over x to the fifth okay so not required if a question instructs you to eliminate negative exponents make that step from here to here the next function here is y equals the root of x plus the cube root of x plus the fourth root of x plus the fifth root of x so in power form this is x to the one half plus x to the one third plus x to the one fourth plus x to the one fifth and these are all going to be addressed each term by the power rule so the first term i get one half x to the negative one half so the one half comes down so one minus one half is going to be negative one half excuse me i have that backwards one half minus one is going to be negative one half the second term here the power of one third comes down i'm gonna take one away so one third take away one that's going to be one take away three-thirds so i get a power of negative two-thirds the third term the one-fourth comes down we'll take away one so one-fourth minus four over four is going to be minus three over four and then the last term will bring the power of one-fifth down take one away so one-fifth take away five-fifths that's going to be negative four-fifths okay and since i'm almost out of room here we'll just leave this in the negative exponent form but if we wanted to eliminate the negative exponents we could write these x terms down in the denominator and then we could convert them to their radical notation but at the end of the day here this is the derivative so in the next example i have the instructions here to evaluate the expression so d over dx of 2 root x plus 3 halves times the cube root of x squared plus 5 4 times the fifth root of x to the fourth power so again the d over dx here is my differential operator it just tells me to find the derivative or to take the derivative of this function inside so at first perhaps we will rewrite this into exponent form so this is going to be two x to the one-half plus three-halves x to the two-thirds so power of two divided by the root three that becomes our exponent now plus five fourths x to the power of four over five so now we have this in form to just apply the power rule so first term the power of one-half comes down so that's going to be two times one-half now x is our variable coming along we have now one-half take away one that's the power rule so that's going to be negative one-half the next term i have the three-halves coefficient coming along power rule i bring down the two-thirds take one away so two-thirds minus three-thirds is going to be negative one-third and then the last term the five-fourths comes along i apply the power rule bring down the power of four-fifths take away one so four-fifths minus five over five is going to be negative one-fifth so you'll notice here that these coefficients are reciprocals and so reciprocals multiply to one so this is just going to be x to the negative one-half plus x to the negative one third plus x to the negative one fifth and so this is going to be 1 over the root of x this will be 1 over the cube root of x and this will be one over the fifth root of x so our derivative simplifies down to this expression in this example i'm given some information about two functions f and g so i'm told that f prime of one is pi and g prime of 1 is e and i'd like to find this expression 2f minus 3g prime of 1. so what we have here is just a demonstration of this property of linearity so this 2f minus 3g prime of 1 is going to be 2 f prime of 1 minus 3 g prime of 1. so that's the property of linearity right this operation of differentiation applies across the summer difference so we get f prime and g prime the operation comes along and we get the derivative of the f and g terms so now i'm told that f prime of 1 is pi and then g prime of 1 is e so this is simply going to be equal to the result 2 pi minus 3 so when you see something like this it's just telling you to apply the operation to this expression 2f minus 3g and since we know the operation of differentiation is linear the derivative jumps onto each of the functions the operation between comes along and then the constants come along and now i have my givens f prime of 1 is pi g prime of 1 is e so notice in this section so far we've not talked about the derivatives involving products or quotients the derivatives of products and quotients require special treatments and they require their own rules in fact so we'll talk about the product rule and the quotient rule in the next section up to this point however if we see a product or a quotient we probably want to to take a step to simplify and then be in a place where we can just apply the power rule so we have to be careful to not naively apply the derivative to products and quotients anytime i see a product or a quotient and i'm looking for the derivative it's going to require the product rule or the quotient rule which we'll talk about in the next section so to this point we could approach something like this by simplifying so i have a quotient here x cubed minus 3x squared minus 2x divided by x squared so what i want to do here is i want to treat this in terms of things that we know how to find the derivative of so what i can do here is i can split this up into three terms x cubed over x squared and then minus 3x squared over x squared and then minus 2x over x squared so this just becomes now the derivative of x minus 3 minus 2 over x all right bye so we just simplify by cancelling canceling terms or applying our exponent laws so the power of three minus two gives us a power of one these two are going to cancel power of one minus two is going to give us the power of negative 1. so this 2 over x i might go ahead and just write 2x to the minus 1. and so now the derivative is going to be very simple the derivative of x is just 1. so the linear term you can just effectively take the coefficient the derivative of negative three is going to be zero and then the last term here we'll apply the power rule the negative one comes down so negative one times negative two will be positive two take one away i'll have one plus two x to the negative two as my derivative and if we wanted to eliminate the negative that could be the same thing as one plus two over x squared so a horizontal tangent line occurs when the slope of the tangent line is zero in other words when the derivative is zero so if i want to find locations where a function has horizontal tangents what i do is i set the derivative equal to 0 and solve for x or solve for whatever variable is involved so horizontal tangents have a deeper significance in applied settings and we'll see this again in chapter 4 when we get to the topic of optimization so if i think about the function f of t equals t cubed minus 27 t plus 5 i'd like to first find the equation of the tangent line to this function f at t equals zero so i need to find the derivative in order to find the slope of the tangent line so we'll have f prime of t by the power rule is just going to be three t squared minus 27 and so at t equals 0 we're going to have the derivative at 0 being equal to negative 27. so the slope of our tangent line is going to be negative 27 now we need our point so when t equals zero f of t or excuse me the y value is going to be f of zero and plugging in zero into the original we're clearly going to get f of t f of zero equals five so our point is zero five so our tangent line our tangent line has the equation y minus y one so that's going to be y minus five equals our slope of negative 27 times x minus zero in this case this simplifies simply to negative 27 x plus five so that's the equation of our tangent line at t equals zero now in the next part i'd like to locate all horizontal tangents so the t values where we have a horizontal tangent so a horizontal tangent means the derivative will equal zero so our derivative being three t squared minus 27 we'll just set that equal to zero and solve so adding 27 we get 3t squared equals 27. divide by three that means t squared has to be nine so solving we're going to get two t values plus n minus three so when t equals three and when t equals negative three the tangent will be horizontal in other words the tangent will have slope 0. a common type of question is to apply the concept of the derivative given a graph so here i have two functions f and g on the graph and i'd like to find the value of this expression the cubed root of g prime of 2 minus f prime of 2. so looking at the graph here if i look here at the value of 2 for our function f we're going to be right here now again this derivative represents the slope of the tangent so the slope of the tangent if i follow it up to the curve here it sure looks to me that the slope of the tangent here at x equals two is going to be zero so i'm going to make the conclusion that f prime of two is equal to zero now for g prime of two so again we're at the value of two i'm going to be up here on the line well the derivative of a linear function this is certainly linear and the derivative of a linear function is going to be its slope so i just want to pick two points and find the slope so this looks like the point 3 3 this looks like the point 0 1 so f prime excuse me g prime of 2 is just going to be the derivative at 2 well this is a linear function so the linear function the derivative is going to be the slope so g prime of 2 in this case is going to be y 2 minus y 1 divided by x 2 minus x 1 and so that's going to be 2 over 3. so our expression here the cubed root of g prime of 2 minus f prime of 2 is going to be equivalent to the cubed root of 2 3 minus 0. so the last derivative rule that we pick up in section 3.3 is going to be the rule for the exponential function with base e and so what we find is that the derivative of e to the x is e to the x and i've shown you a justification here through the limit definition so now we have an additional rule to add to our list the derivative of e to the x is going to be e to the x so be careful it can be tempting to mix up the power rule and the exponential rule and say something like this the derivative of e to the x is apply the power rule and get x times e to the x minus 1. but these two rules address two different situations one has the variable in the exponent and that's the exponential function the other is going to have the variable in the base and the exponent is going to hold a number and that's going to be the power rule so when i have variable and base number in the exponent that's a power that's the power rule when i have the variable being in the exponent and the number in the base that's an exponential function which requires the exponential rule so just take care to not mix these two rules together so let's find a couple more derivatives with this new rule that we have here's the function h of t is equal to the fourth root of t minus four e to the t plus e times t to the fourth power so this is going to be the same thing as t to the one fourth and then minus four e to the t plus e times t to the fourth power so the derivative first term is just going to require the power rule so we have 1 4 comes down and then now we're going to take one away so that's going to be 1 4 minus 1 which is going to be negative 3 4. the second term is minus four e to the t the four comes along the derivative of e to the t is e to the t and now the last term here we have e times t to the fourth power well here e is just acting as a constant so we're gonna get 4 e t take 1 away cubed the next function here is k of x equals e to the x minus x to the e plus e times x plus e so the derivative term by term the derivative of e to the x is e to the x now the second term here is x to the e so i have variable raised to a number so this is going to be the power rule so i'm going to get bring the power down take one away so i get e x to the e minus 1. the third term here is e times x this is just a linear term so for a linear term we just get the coefficient so i get plus e and then the last term here is just a constant the derivative of the constant is going to be zero so to wrap things up in this section we'll talk briefly about higher order derivatives so the idea here is very simple if f is a differentiable function then f prime may also be differentiable and so what we would say is that the derivative of f prime is called the second derivative so we'll denote this by the prime notation we say f double prime we also have the leibniz notation here d squared y over dx squared which represents the second derivative as well so we can continue this it could be the case that the second derivative is differentiable in which case i can differentiate to find the third derivative so we have f triple prime we may also write this notation f with three in parentheses written above as a superscript and we also have the leibniz notation as well these all represent the third derivative and so in general beyond the third derivative we'll use this notation so for instance the fourth derivative we would write f with the four in parentheses sort of written as a superscript and then we have the leibniz notation for the nth derivative and beyond so a couple of functions here i want to find f prime f double prime and f triple prime in other words the first second and third derivatives for these given functions so the first derivative of this function in the first part of this example we'll just apply the power rule so that's going to be 5x to the fourth minus 6x squared so there's the first derivative now we can write the second derivative so f double prime apply the power rule that's going to be 20 x cubed minus 12 x and then we can write the third derivative this will be f triple prime the third derivative is going to be 60x squared and then the derivative of negative 12x is just going to be minus 12. so there's our first second and third derivatives for this function this fifth degree polynomial so this could continue we could write a fourth derivative we could write a fifth derivative and so on so the second function here is the function f of x equals 2 e to the x so the first derivative is going to be 2e to the x what we'll see is that the second derivative will also be 2e to the x and we're also going to see the third derivative is going to be equal to 2 e to the x and in fact every derivative of 2 e to the x this can continue to infinity every derivative even the hundredth or the 200th derivative is going to be 2e to the x so this is the idea of higher order derivatives in calculus we have a lot of use with the first and second derivatives typically in application we don't see much use with anything beyond the second derivative although we can calculate third fourth fifth and so on derivatives if we would like to if the function remains differentiable to that point