Using everything that we've learned about finding derivatives so far, we know how to find the derivative of something like \(4x + 5\). We also know how to find the derivative of some variable like \(x^3\). But what if we put those 2 together and we're asked to find the derivative of \((4x+5)^3\)? How could we go about that? Well, we could go ahead and multiply out that \(4x+5\) 3 times so that we could then just use the power rule.
But that sounds a little bit messy and time-consuming, and there's actually a much quicker and easier way to find this derivative using what's called the chain rule. So, here, I'm going to show you exactly what the chain rule is and how to use it, and we'll work through some examples together. So let's go ahead and get started. Now, the chain rule tells us that in order to find the derivative of a composite function, that's one function inside of another function, we want to start from the outside and work our way inside. So what exactly do I mean by that?
Well, coming down here to our chain rule when finding the derivative \( \frac{d}{dx} \) of \( f(g(x)) \), that's one function inside of another function, I'm going to start by finding the derivative of that outside function \( f \). So I want to find \( f' \). When I find the derivative of that outside function, I'm going to treat that inside function as though it's just a variable and completely leave it as is. So this first term of my chain rule is \( f'(g(x)) \), the derivative of that outside function, treating that inside function as though it's just a variable. So what does that look like for an actual function?
Well, coming back over here to our example, \( \frac{d}{dx} \), our derivative of \((4x+5)^3\). Here, I see that on the outside, I have this stuff being cubed. And on the inside, I have this \(4x+5\). So starting out with my chain rule here, I want to find the derivative first of that outside function. Now, my outside function is this stuff being cubed.
And I know that when dealing with something cubed, I can just use the power rule to find its derivative by pulling that 3 out to the front, multiplying by it, and decreasing that power by 1. So this first term is 3 times that inside function, treating it like a variable, leaving it completely alone to the power of \(3 - 1\) or \(2\), having used the power rule. Now for our second term here, that's just the first term of our chain rule, we said we wanted to work our way inside. So we now want to find the derivative of that inside function, \( g(x) \), and multiply by it. So we want to multiply here by \( g'(x) \).
Now doing that to our actual function here, our inside function is \(4x+5\). The derivative of \(4x+5\) is just 4. So I want to multiply by 4 and this is my full derivative. Now I can do a little bit of simplification here by multiplying my 3 and my 4 to give me \(12(4x+5)^2\). And that's my full derivative, having used the chain rule here to find \( f'(g(x)) \times g'(x) \). Now you may also see the chain rule written in a different notation referred to as the Leibniz notation that we see here.
Now in this notation, they're referring to \( y \) as your outside function and \( u \) as your inside function. So that tells us \( \frac{dy}{dx} \), that's the derivative of my whole function, is equal to \( \frac{dy}{du} \), that represents my first term here, \( f'(g(x)) \times \frac{du}{dx} \). That's that second term, \( g'(x) \). So if you see this notation, just know it means the same exact thing. Now let's work through one more example together here.
We want to find \( f'(x) \) here using the chain rule for this function, \( f(x) = 2(3x^2 - x)^4 \). Now the first thing that we want to do here is identify what our inside and outside functions are. Now, usually, it's pretty easy to identify that inside function because it will most often be contained in parentheses. Like, here, we have \(3x^2 - x\). That makes our outside function \(2\) times all this stuff to the power of \(4\).
Now applying our chain rule here to find \( f'(x) \), our derivative, we want to start from the outside and find the derivative of \(2\) times the stuff to the power of \(4\). Now we can do so here using the power rule, multiplying it by 4, pulling that out front, and decreasing that power by 1. This gives us \(8(3x^2 - x)^3\). Then from here, we want to multiply by the derivative of that inside function. Now here, the derivative of \(3x^2 - x\) using the power rule here is going to give me \(6x - 1\), making sure we're taking the derivative of both of these terms.
And this is my final derivative here, \( f'(x) \). I started by taking the derivative of that outside function, treating that inside function like a variable, and then multiply that by the derivative of that inside function. Now remember that the most important part with all of these rules for derivatives is getting a ton of practice. So let's keep practicing the chain rule coming up next. I'll see you there.