Proof: the derivative of ln(x) is 1/x | Advanced derivatives | AP Calculus AB | Khan Academy

What we're going to do in this video is prove to ourselves that the derivative with respect to X of natural log of x is indeed equal to 1/x.

So let's get started. Just using the definition of a derivative, if I were to say the derivative with respect to X of natural log of x, that is going to be the limit as Delta X approaches zero of the natural log of x + Delta x minus the natural log of x, all of that over Delta X.

Now, we can use a few logarithm properties. We know that if I have the natural log of A minus the natural log of B, this is equal to the natural log of A over B. So we can use that right over here where we have the natural log of something minus the natural log of something else.

So all of this is going to be equal to the limit as Delta X approaches zero of the natural log of this divided by that: so (X + Delta x)/X. All of that over Delta X. Actually, let me just write it that way: all of that over Delta X.

Once again, all I did was say if I have natural log of that minus natural log of that, that's the same thing as natural log of this first expression divided by that second expression. It comes straight out of our logarithm properties.

Well, inside of this logarithm, x/X is 1, and then Delta x/X we could just write that as Delta X/X. So that's another way of writing that. We could put a 1 over Delta X out front, so we could say this is the same thing as this is equal to the limit as Delta X approaches zero of, I'll do this in another color, so this I can rewrite as 1 over Delta x times the natural log of (1 + Delta X/X).

Let me close that parenthesis. Now, we can use another exponent property. If I have a * the natural log of B, that is equivalent to the natural log of B to the A, and so here this would be the A in that case. So, I could bring that out and make that an exponent on this.

So this is all going to be equal to the limit as Delta X approaches zero of the natural log of (1 + Delta x/X) to the (1/Delta X) power. Now, this might start to look familiar to you; it might start to look close to the definition of e, and we are indeed getting close to that.

In order to get there fully, I'm going to do a change of variable. I am going to say: let's let N equal Delta X/X. Well, in that case then, if you multiply both sides by X, you get Delta X is equal to NX. Again, just multiplied both sides of this equation by X and swapped the sides there.

Then, if you wanted 1 over Delta x, that would be equal to 1 over NX, which we could also write as 1 over n times 1/x. Actually, let me write it this way; actually, that's the way I do want to write it.

So these are all of the substitutions that I want to do for my change of variable. We also want to say, well look, as Delta X is approaching zero, what is n going to approach? Well, as Delta X approaches zero, we have n will approach 0 over X. Well, that's just going to be zero for any X that is not equal to zero.

And that's okay because zero is not even in the domain of the natural log of x, so this is going to be, for our domain, it works that as Delta X approaches zero, n approaches zero. You could think about it the other way around: as n approaches zero, Delta X approaches zero.

Now let's do our change of variable. So if we make the substitutions, instead of taking the limit as Delta X approaches zero, we are now going to take the limit as n approaches zero of the natural log of, give myself some parentheses, and I'll say 1 + n, and then all of that is going to be raised to the (1/n * 1/x).

That's what 1 over Delta X is equal to. This is 1 over Delta X right over here, which we have over here, and it's the same thing as 1/n * 1/x.

So let me write that down: so this is the same thing as 1/n * 1/x. Now we can use this same exponent property to go the other way around. Well, actually, let me just rewrite this another time.

So this is going to be the same thing as the limit as n approaches zero of the natural log of 1 + n to the (1/n). If I raise something to an exponent, and that's times something else, that's the same thing as raising it to the first exponent and then raising that to the second value.

This comes once again straight out of our exponent properties. Now we can use this property the other way to bring this 1/x out front, but in fact, the 1/x itself is not affected as n approaches zero.

So we can even take it completely out of the limit. So we could take it all the way over there, and this is when you should be getting excited.

This will be equal to (1/x) times the limit as n approaches zero of the natural log of (1 + n) to the (1/n). Let me do that in orange: (1 + n) to the (1/n).

Now, what really gets affected is what's going on inside of the natural log; that's where all of the n's are. So let's bring the limit inside of that.

So this is all going to be equal to, get myself some space, a little bit of extra chalkboard space, this is going to be equal to (1/x) times the natural log of the limit as n approaches zero of (1 + n) to the (1/n).

Close those parentheses. Now, this is exciting! What is inside the natural log here? Well, this business right over here: this is a definition of the number e.

So that is equal to e. Well, what's the natural log of e? Well, that's just one. So it's (1/x) * 1. Well, that is indeed equal to (1/x), which is exactly the result that we were looking for: that the derivative with respect to X of natural log of x is (1/x). Very exciting!