Implicit differentiation, product and chain rules at once

Let's say Y is equal to the natural log of x to the X power. What we want to do is we want to find the derivative of Y with respect to X. So I encourage you to pause this video and see if you could do it.

So when you first try to tackle this, this is a little bit daunting. We know how to take the derivative of constants to some X power, but how do we take a derivative of some type of a function—in this case, the natural log of X to the X power? The answer here is to use some of our logarithmic properties, and then we're going to do a little bit of implicit differentiation.

So the first thing that we want to do—and actually, let me rewrite this with a little bit of space—so this was the natural log of X to the X power. The first thing I want to do is get rid of this X as an exponential, and I want to be able to apply the product rule somehow. The way we're going to do that is by taking the natural log of both sides.

So, take the natural log of both sides, and you might say, "Well, why is that helpful?" Well, if I'm taking the natural log of something to an exponent, this is the same thing—actually, let me write this down as a property that you may or may not remember from your logarithmic properties.

So if I have, I could write log—or I'll just write natural log of—if I have natural log of a to the B power, this is the same thing as B times the natural log of a. That's just a standard logarithmic property. So by taking the natural log of both sides, this exponent can now become out front and scale the natural log function.

Now we can bring that out front, and let's just rewrite everything. So we get the natural log of Y is equal to—let me put that in parenthesis—so the natural log of Y is equal to X times the natural log of X.

So there you have it. By just taking the natural log of both sides and using this logarithmic property, we were able to get that. Now you're saying, "Well, gee, how is this actually going to be useful for us?" Well, now we can implicitly take the derivative of both sides of this.

Actually, let me scoot this over to the right a little bit just so that I can have space for my derivative operator. So there you go, scooted that over. And so now, let's take the derivative with respect to X of both sides.

So let me take the derivative with respect to X of the left-hand side and the right-hand side. Now on the left-hand side, this is going to be essentially an application of the chain rule. When you learn implicit differentiation, it's really just application of the chain rule.

It's the derivative of the outside function with respect to the inside function. So the natural log of Y with respect to Y, the derivative of that is just going to be 1 over Y. That is going to be equal to—well, this is going to get interesting a little bit.

Actually, let's do some stuff on the side a little bit just—well, let me just—the first thing we want to do here is apply the product rule. It's the derivative of the first expression, so it's just going to be 1 times the second, I guess you'd say, function—so times the natural log of X, plus the first function, just X, times the derivative of the second function.

What's the derivative of the natural log of the natural log of X? Let's do that separately. So if I am trying to take the derivative with respect to X of the natural log of the natural log of X—well, here again, I can apply the chain rule. The derivative of that magenta function with respect to the inside function is going to be 1 over the natural log of X, and then times the derivative of the inside function with respect to X, so times 1 over X.

So this is equal to 1 over X times the natural log of X. So the derivative of this second function right over here is 1 over X times the natural log of X. Let's see—that X cancels out, and so we are left with—I'll just write all of this in this blue color—so 1 over Y times the derivative of Y with respect to X is equal to the natural log of the natural log of X plus 1 over the natural log of X.

Now to solve for the derivative, we can multiply both sides by Y. So let's do that. So we're going to multiply that side by Y, and we're going to multiply this side times Y. What are we going to get? Well, on the left-hand side—that's why we multiplied by Y—we just have the derivative of Y with respect to X.

The derivative of Y with respect to X is equal to, well, Y—our original thing that we had—Y was equal to the natural log of X to the X power. So we’re essentially multiplying both sides times the natural log of X to the X power.

So this is going to get a little bit messy here. We could write it the way I wrote it just now without it being distributed. Actually, let me just leave it like that. So it's going to be—and so we deserve our drum roll right now because this is quite involved—the natural log of the natural log of X plus 1 over the natural log of X—all of that times the natural log of X to the X power.

So that was quite involved. If someone said, "Well, what is the derivative of Y when X is equal to E?"—if someone says, "What is this equal to when X is equal to E?"—well, we could evaluate this when X is equal to E. This would be the—and I know I just made that up just now—so if, like, the original question wasn't just what is dy/dx, if they said, "What is dy/dx when X is equal to E?"—if that was the original question, then we could evaluate it.

So where we just replace all of these with E's—there'd be an E there, an E there, an E there, and an E there—and I just picked the value E because it's easy to evaluate. So the natural log of E is 1, natural log of 1 E to the zero power is 1, so all of that just becomes zero. The natural log of E is 1, so this whole expression right over here becomes 0 plus 1 over 1, so it just becomes 1.

Then the natural log of E is the natural log of E is 1, and you're going to have 1 to the eth power—you could raise 1 to any power and you're just going to get 1.

So it's 1 times 1 is equal to 1, so I just thought it would be fun to try to evaluate that at a value that would be somewhat clean.