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## Differential Calculus

### Course: Differential Calculus>Unit 3

Lesson 4: Implicit differentiation (advanced examples)

# Derivative of ln(x) from derivative of 𝑒ˣ and implicit differentiation

How can you find the derivative of ln(x) by viewing it as the inverse of e^x? Created by Sal Khan.

## Video transcript

- [Voiceover] We know that the derivative with respect to x,of e to the x, is equal to e to the x, which I think is one of the neatest things in mathematics. It just builds my respect for the number e. But that's not what we're here to do, to just praise e. What we really want to think about is, what's the derivative of the inverse function? What's the derivative, with respect to x, of the natural log of x? We've done this several times, where we know the derivative of a function and we now want to find the derivative of the inverse function. What we can do, let's say y is equal to the natural log of x. This is another way of saying that y is the power that we raise e to, to get to x. So, this is equivalent to saying that e to the y power is equal to x. Now we can take the derivative of both sides of this equation with respect to x. So let's do that. The derivative of both sides with respect to x, do a little bit of implicit differentiation. Really just an application of the chain rule. So, on the left-hand side right over here, this is going to be the derivative of e to the y with respect to y, which is just going to be e to the y times the derivative of y with respect to x. On the right hand side here, the derivative of x with respect to x is one. Now, to solve for the derivative, we just divide both sides by e to the y. We get the derivative of y with respect to x is equal to one over e to the y power. Now, what is y equal to? Well, we know that y is equal to the natural log of x. So let's write that down. This is equal to one over e to the natural log of x. What is e to the natural log of x going to be? The natural log of x is the power I need to raise e to, to get to x. So if I actually raise e to that power, to that exponent, I'm going to get x. This is going to be equal to one over x. So this thing simplifies to x. We are done. We just figured out if y is equal to the natural log of x, the derivative of y with respect to x is one over x. Or you could say the derivative of the natural log of x with respect to x is equal to one over x. So this is equal to, this right over here, is equal to one over x. One over x, which is also a pretty neat result in mathematics. Not quite as exciting as this one, but still pretty neat.