Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Exponential functions. The inverse of the natural log of X is E to the X. So for every real number, XE to the X equals the inverse of the natural log of X. Or sometimes that's written as EXPX, the inverse equation for E to the X&L&X. So composition an inverse put into the original function should undo itself. So E to the LNX equals X when X is greater than 0. Because we have to have the natural log of only positive numbers, and the natural log of E to the X = X for all X's. And the reason that one's true is E to some power, no matter what that power is, is already a positive number. So the natural log of a positive number is just the X. So A to the X is going to equal E to the natural log of A to the X because that E and that LNX would or that E and that Ln would just cancel. So then by using the power rule, we could bring the X down in front and we get EX to the LNA. So A to the X = e X to the LNA. Now we can use some of the rules for natural logarithms on a base of a different sort of a any constant. So E to the X 1 * e to the X2. When the bases are the same, we just add the exponents. E to a negative exponent means one over. If we have the same base and we're dividing, we subtract the exponents and powers to powers mean we multiply. Just a quick review there of some of our exponent rules. The exponent exponential function is differentiable because it is the inverse of a differentiable function whose derivative is never 0. So if we have F of X equaling LNX, we know that F prime of X or the 1st derivative is 1 / X. So F inverse prime of X is equal to 1 / 1. But now we need to actually figure out what the inverse function is. So come on over here. If we switch our X and our Y and solve for Y, we'd get X equal L&Y. To get rid of that natural log, we take and put it in the exponential form. So E to the X equal Y. So the inverse function says E to the X. So when we find the derivative of the inverse function, it's 1 divided by the derivative of the original function having the inverse inputted into that first derivative. So 1 / 1 / e to the X is E to the X. This is a really powerful concept. It's the first equation that the derivative is the same thing as the original function. So the derivative of E to the X is just E to the X. So an example, if we have Y equal E to the -5 XY prime is just E to the -5 X. Now that -5 X is is a function in terms of X. So we're going to multiply it by the chain rule times the derivative of a negative 5X and we're going to get -5 E to the -5 X. We have a theorem that states the number E is a limit E equal the limit as X approaches 0 of 1 + X to the 1 / X. And our proof for this is if we let F of X = l and XF prime of X is 1 / X. So F of one is 0 and F prime of 1 equal 1. So if we had stuck one in here, 1 / 1 is one. We're going to use all those pieces and the official definition of a derivative. So we have the limit as H goes to 0 of F of X + H -, F of X over HF prime of one. Then is going to equal the limit as H goes to 0 of F of 1 + H -, F of one all over H Well, the natural log of 1 + H -, F of one we found up here is 0. So now if we think of the limit as H goes to 0 of 1 / H natural log of 1 + H, we're going to take this 1 / H up by the power rule. And the limit as H goes to zero of the natural log of 1 + H to the one over H. Now F prime of one way back here we said was 1. So one has to equal all of that by the transitive property. If that's the case. Now we're going to take this and we're going to put it into. We're going to take E on each side, so E to the first. The E and the natural log of all this are going to cancel, leaving us just the limit as H goes to 0 of 1 + H to the one over H. This is also true if K equal 1 / H So whenever we see 1 / H, we're going to put in K, and if we see H, that's really 1 / K. Now if H was going to 01 over something really really small is really really big. So E equal limit as K goes to Infinity of the quantity 1 + 1 / K to the K power. So the derivative of E to some function is just going to be E to the function times the derivative of UDX. The derivative in terms of E to the X is just E to the X because the derivative of X is 1. So we know that the integral of E to the UDU is just going to be E to EU + C Let's look at a few examples. If we have this example, we're trying to find the derivative. So we're given Y equal 1 + 2 xe to the -2 X. We're going to have to use the product rule because we have two different terms multiplied together that each have an X. So we're going to have 1 + 2 X the first term times the derivative of the second, the derivative of the second E to the -2 X times the derivative of that -2 X which is -2 plus the derivative of this first one. Well, the derivative of one is 0 and the derivative of 2X is just two times the second term. If we distribute this out and combined like terms, we can see it simplifies to -4 xe to the -2 X. If we look at another example, this time we have Y equal E sine T of the natural log of t ^2 + 1. Once again the product rows. So the derivative of the first one E sine T, the derivative of sine T because of that's being a function all times the second term plus the derivative of the second term. Well, we've got to use the chain rule here. So the derivative of natural log is 1 / t ^2. But the derivative of t ^2 we also have to find and that's going to get multiplied by the 1st function. If we keep going, we're going to have Y prime equal E sine T. The derivative of sine T is cosine T. All that times the natural log of t ^2 + 1 and then 1 / t ^2. The derivative of t ^2 is 2 TE to the sine T. Simplify that out and that would be our response to that, that example or our answer. Another one, what if we want the derivative of an integral? Well, the derivative of an integral basically by the fundamental theorem of calculus, undo each other. So Y equal Y is going to turn into Y prime. So we're going to put in the upper bound for our unknown T and then we're going to multiply by the derivative of the upper bound minus. We're going to stick in the lower bound and times it by the derivative of the lower bound. So sine E to the lane X. Well, E to the lane is going to reduce to give us just X. So we're going to end up with sine X. The derivative of lane X is 1 / X. The derivative down here is 0. The derivative of any constant is 0. So Y prime is just sine X / X. What if we were given an integral and we want to evaluate what the original function is? So if we know that Y prime equals this, we're going to take and we're going to say if we have E to the 3X, we need to find the derivative of that exponent. And instead of multiplying, we're trying to undo it. Now we're going to divide. So the derivative of 3X is 3, so E to the three X / 3 + 5 E to the negative X. The derivative of negative X is -1, so it's going to be divided by -1. Or we could think about just bringing that out in front and then because it's not bounded, we're going to have a + C there at the end. Thank you and have a wonderful day.