 So previously we've seen that the derivative of e to the x with respect to x is equal to e to the x That is the derivative of the natural exponential is itself What about other exponential functions? What if we want to find the derivative of 2 to the x dx? Is that equal to 2 to the x as well? Well, if you go back to the proof of this statement the squeeze theorem argument We used to derive that the derivative of e to the x itself was dependent upon e to the x That is we derived some inequalities for which those inequalities came from it being e So we shouldn't expect that other exponentials will be their own derivative something special about the number e But what can we do? It turns out that if we use the chain rule We can actually combine what we know about the derivative e to the x and apply that to the derivative of any exponential function So if we want to so what we're gonna prove right now is the derivative of a to the x with respect to x or a is any exponential base, it's just a positive number We get that that's going to equal a to the x times the natural log of a right here So before we proceed to do the proof I want you to kind of examine this for a second Okay, the derivative of a to the x is a to the x times the natural log of a if we think about that in terms of e to the x After all e is a number that's positive So it applies here we take the derivative there if you use this formula This would tell us the derivative is e to the x times the natural log of e Which the natural log of e since natural log is base e this is just equal to 1 you get e to the x times 1 So we we can reproduce that e to the x is a special case of this And so the way I like to think of it is this right here is some type of tariff that has to be paid So that in terms of calculus the the currency should be base e But if we are using a different currency like base 2 base 7 whatever we can still take the derivative But there's sort of a transaction via tariff that we have to pay In order to calculate the derivative of an arbitrary exponential function So how are we going to prove this with a chain rule because chain rule is about function composition f compose with g of x What's the composition and play right here? So it takes a little bit of effort, but notice it right here if I take a to the x This is the same thing as e to the natural log of a to the x I want you to think about for a second if I take a to the x and I add to a 2 and I subtract from it Most we'll have no arguments there that oh the plus 2 minus 2 cancel out So it's just a to the x on the other hand What if I take like a to the x and I times it by 2 and I divide it by 2 again? Not a lot of conflict there if you divide things if you times that multiply by the same number just cancels out Well, what we're seeing right here is that same basic idea here We're gonna Exponentiate by e and then we're going to log Rise is that even we're gonna logarithm. I'm using it as a verb right now We're gonna logarithm by e for which these are inverse operations Exponentiate by e and logarithm by e are inverse operations. So the composition of these two does nothing So that's why equality follows right here But then the benefit of introducing the logarithm is that by logarithmic laws We can bring this fact or this exponent of x out the front as a factor And so now to calculate the derivative of a to the x it equates to taking the derivative of e to the x times the natural log of x Which we'll see is to our advantage to do so. So why is that? Well, the derivative of a to the x with respect to x like I said a moment ago It's the same as taking the derivative of e to the x plus the natural log of a now I want you to focus on that exponent for a moment. Look at that x times the natural log of a Writing this in the slightly larger font on the screen right here We have composition of functions now The exponent of the exponential is this inner function x times the natural log of a and then the outer function is going to be this Exponential function for which by the chain rule We're gonna take the outer derivative right here e to the x times the natural log of a power and then we have to take The inner derivative right here, which we're taking the derivative of x times the natural log of a now here Remember, we're taking the derivative with respect to x x is the variable that's changing the base a is fixed in this consideration And so if you take the derivative like a 2x we know that is that's just gonna be a 2 if you take the derivative of 5x That's just gonna be a 5 the same thing also applies here The natural log of a is just a number like 2 or 5 sure It's probably an irrational number, but it's still a number. It's just the coefficient of x So when you take the derivative of the natural log of a times x the derivative turns out to be the natural log of a and Which case then when you look at the outer derivative Wait a second the outer derivative is just this guy right here, which this one remember was a to the x and so you make that substitution and we get a to the x times the natural log of a and So this that allows us that if we want to take the derivative of say like 2 to the x with respect to x This was to be 2 to the x times the natural log of 2 It's just like e to the x you get back itself, but then you also have to pay the tariff because you weren't working base e How about this one if we take the derivative of s which is e to the t well the derivative of S with respect to t you're taking the derivative e to the t right here Where you're gonna just get e to the t times the natural log of 3 again remember to pay the tariff That's the most common thing that students mess up if they take a derivative of an exponential other than base e You sometimes forget to pay the tariff All right, how about about this one this time s equals 8 times 10 to the 1 over t power So if you want to calculate the derivative of s with respect to t we have to take the derivative 8 times 10 to the 1 over t power Well since 8 is just a constant multiple we can bring it out and so we're gonna get 8 times 10 to the 1 over t power and Much like we saw in the proof of this thing. There's a chain rule going on here There's gonna be the inner function of 1 over t and then there's the outer function Which is 10 to the power of something Expedition base 10 and so when we calculate the derivative here We're going to get our outer derivative which will be 10 to the 1 over t Times the natural log of 10 you have to pay the tariff with the outer derivative and then the inner derivative Well, we have to take the derivative of 1 over t Which you might want to think of 1 over t as t to the negative 1 power That can be very useful here. So then the power rule applies So we're gonna get 8 times 10 to the 1 over t times the natural log of 10 Next when we take the derivative of t to negative 1 we're getting negative t to the negative 2 and so putting this all together in a slightly More condensed simplified format. We're gonna get Negative 8 times the natural log of 10 because those are just the numbers I'm gonna put in front through the coefficients We're gonna times that by 10 to the 1 over t power and this sits above t squared right there Thus giving us the derivative of this function. So we can combine these Exponential functions with the other functions and derivative rules we've developed so far We can take the derivative of any exponential base Just make sure that you pay the tariff when you are not working base e