Worked example: Derivative of ln(ÃÂx) using the chain rule | AP Calculus AB | Khan Academy
So we have here F of x being equal to the natural log of the square root of x. What we want to do in this video is find the derivative of F. The key here is to recognize that F can actually be viewed as a composition of two functions, and we can diagram that out.
What's going on here? Well, if you input an x into our function f, what's the first thing that you do? Well, you take the square root of it. So if we start off with some x and you input it, the first thing that you do is you take the square root of it. You are going to take the square root of the input to produce the square root of x. Then what do you do? Well, you take the square root and then you take the natural log of that. So then you take the natural log of that.
You could view that as inputting into another function that takes the natural log of whatever is inputted in. I’m making these little squares to show what you do with the input. Then what do you produce? Well, you produce the natural log of the square root of x, the natural log of the square root of x, which is equal to F of x. You could view f of x as this entire set or this entire, I guess you could say, this combination of functions right over there. That is f of x, which is essentially a composition of two functions. You’re inputting into one function, then taking that output and inputting it into another.
So you could have a function U here which takes the square root of whatever it’s input in. So U of x is equal to the square root of x. Then you take that output and input it into another function that we could call V. What does V do? Well, it takes the natural log of whatever the input is. In this case, in the case of f or in the case of how I've just diagrammed it, V is taking the natural log. The input happens to be the square root of x. So it outputs the natural log of the square root of x.
If we wanted to write V with x as an input, we would just say, well, that’s just a natural log of x. As you can see here, F of x, and I color-coded it ahead of time, is equal to F of x is equal to the natural log of the square root of x. So that is V of the square root of x or V of U of x. So it is a composition, which tells you that, okay, if I'm trying to find the derivative here, the chain rule is going to be very, very, very, very useful.
The chain rule tells us that F prime of x is going to be equal to the derivative of, you can view it as the outside function with respect to this inside function. So it’s going to be V prime of U of x times the derivative of this inside function with respect to x. So that's just U prime of x.
How do we evaluate these things? Well, we know how to take the derivative of U of x and V of x. U prime of x here is going to be equal to, well, remember, the square root of x is just the same thing as x to the 1/2 power. So we can use the power rule. Bring the 1/2 out front. So it becomes 1/2 x to the, and then take off one of that exponent. So that’s 1/2 - 1 is -1/2 power.
And what is V of x, sorry, what is V prime of x? Well, the derivative of the natural log of x is 1/x. We showed that in other videos. So we now know what U prime of x is. We know what V prime of x is, but what is V prime of U of x? Well, V prime of U of x, wherever we see the x, we replace it. Let me write that a little bit neater. We replace that with U of x.
So V prime of U of x is going to be equal to, is going to be equal to 1 over U of x, which is equal to, which is equal to 1 over U of x is just the square root of x. One over the square root of x. So this thing right over here we have figured out is 1 over the square root of x. And this thing, U prime of x, we figured out is 1/2 x to the -1/2. I could rewrite that as 1/2 * 1 over x to the 1/2, which is the same thing as 1/2 * 1 over the square root of x, or I could write that as 1 over 2 square roots of x.
So what is this thing going to be? Well, this is going to be equal to, in green, V prime of U of x is 1 over the square root of x times U prime of x is 1 over 2 * the square root of x. Now, what is this going to be equal to? Well, this is going to be equal to, this is just algebra at this point: 1 over 2 and the square root of x times x is just x. So it just simplifies to 1/2 x.
So hopefully this made sense, and I intentionally diagrammed it out so you start to get that muscle in your brain going of recognizing the composite functions and then making a little bit more sense of some of these expressions of the chain rule that you might see in your calculus class or in your calculus textbook. But as you get more practice, you'll start to— you'll be able to do it essentially without having to write out all of this. You’ll say, okay look, I have a composition: this is the natural log of the square root of x. This is V of U of x.
So what I want to do is I want to take the derivative of this outside function with respect to this inside function. So the derivative of the natural log of something with respect to that something is 1 over that something. So it is 1 over that something. The derivative of the natural log of something with respect to that something is 1 over that something.
So that’s what we just did here. One way to think about it: what would natural log of x be? Well, that would be 1/x. But it's not the natural log of x; it’s 1 over the square root of x. So it’s going to be 1 over the square root of x. So you take the derivative of the outside function with respect to the inside one, and then you multiply that times just the derivative of the inside function with respect to x and we are done.