The Chain Rule - Online Tutor, Practice Problems & Exam Prep

Intro to the Chain Rule

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.

Find the derivative of the function.

$f(x)=\sqrt{(5x^2-3x)}$

• A. $\frac12\sqrt{10x-3}$
• B. $\frac{1}{2\sqrt{10x-3}}$
• C. $\frac{10x-3}{2\sqrt{5x^2-3x}}$
• D. $\frac{1}{2\sqrt{5x^2-3}}$

Find the derivative of the function.

$y=(8x^3-2x{)}^{3/2}$

• A. $\frac{3}{2}\sqrt{24x^2-2}$
• B. $(36x^2-3)\sqrt{8x^3-2x}$
• C. $\frac{3}{2\sqrt{24x^2-2}}$
• D. $(12x^2-1)\sqrt{8x^3-2x}$

Find the derivative of the function.

$f\left(t\right)=(3t^2+7t-2{)}^{10}$

• A. $10(6t+7)^9$
• B. $9(6t+9) (3t^2+7t-2)^9$
• C. $(6t+7)(3t^2+7t-2{)}^9$
• D. $10(6t+7)(3t^2+7t-2{)}^9$

Intro to the Chain Rule Example 1

In this example problem, we're asked to find dy/dx, our derivative for the function y=2x−1(4)×3+x2. Now I see a couple of different things happening here. Looking at this first term, I see one function inside of another function, and then this is multiplied by another function inside of another function. So that tells us that we're going to have to combine a bunch of the rules that we've already learned. This can get a little messy, so it's important here to stay organized and know what step you're on and not leave out any terms.

Because we have functions inside functions and they're also multiplying, that tells us that we're going to be using both the chain rule and the product rule. So let's keep that in mind as we work through this derivative. Now here we're finding our derivative dy/dx. So let's start here by first looking at these terms altogether. We have to use the product rule, these are being multiplied.

So what does that tell us here? Well, remember that when working with the product rule, we're going to take that left function, leave it as is. So I have 2x−1(4). Then, we're going to multiply that by the derivative of the function on the right. So that's a ddx of 3+x2. Now I haven't applied my chain rule yet, but when we take this derivative, we are going to use the chain rule. Now I'm going to add this together, still using my product rule here, times that right side function as is 3+x2, keeping that as is, again, multiplying by the derivative of our other function, this time our left function. Remember, our product rule is always one original function times the derivative of the other function. So this here is the derivative of 2x−1(4). Again, here taking this derivative, we are going to have to use the chain rule.

When using the product rule, big picture of this function involves two functions being multiplied together. Now diving into each of these individual terms, we want to actually take these derivatives. So here, we want to stay super organized, keeping all of the terms that we need to, not dropping anything off or forgetting anything. So remember, this is all being multiplied by 2x−1(4).

Now starting with that outside function and then working our way in, using the power rule for that outside function, we have 2 times that inside function, 3+x. All of that is raised to the power 2-1 because we use the product rule here. 2-1 is just 1, so it doesn't need to be written there.

Then moving on to that inside function, the derivative of 3+x. Now the derivative of a constant is just 0 and the derivative of x is just 1. So this is really just all being multiplied by 1, which ultimately won't do anything and can eventually drop off that 1. But we want to stay organized and I'm going to write everything that we need to so that you can see exactly what's happening in this problem.

So we have fully taken this first term and taken the derivatives that we need to. Let's move on to our second term here from our product rule. So here I'm going to add this together with 3+x2. Don't forget that that's right there before moving on to taking our last derivative here again using the chain rule. Let's identify our inside function here which is 2x−1, and, of course, our outside function, which is all of that stuff raised to the power of 4.

Now using our chain rule here, I'm taking this derivative. Starting from the outside, working our way in, our outside function is that to the 4 power. So we're going to pull that 4 out to the front using the power rule and then subtract 1. So this becomes 4 times that stuff in parentheses. We're just treating it like a variable at this point, all raised to the power of 4-1, which is just 3.

Then applying our chain rule, we're not done yet. Don't forget this next step here, multiplying by the derivative of that inside function we started from the outside, we worked our way inside. Now the derivative of that inside function, 2x−1, the derivative of that inside function is just 2. So this is being multiplied by 2,

Intro to the Chain Rule Example 2

In this example problem, we're asked to find the derivative of the function \( y = \frac{(2x - 1)^4}{3 + x^2} \). Now, looking at this problem, it might seem a little bit intimidating. You should go ahead and try this out for yourself before checking back in with me if you get stuck anywhere or if you just want to check if you got the correct answer. But I'm going to go ahead and break this problem down for you fully step by step. So let's jump in here. Now, in finding our derivative for this function, I recognize that there are a couple of different things happening here.

There is division happening between these two functions, which tells me that I need to use the quotient rule. But within my numerator and my denominator, I have a function inside of a function for both of these, telling me that at some point I'm going to need to use the chain rule. So, let's break this down. Now, in finding my derivative \( \frac{dy}{dx} \), we know we need to use the quotient rule, and the quotient rule tells us \( \text{low} \times \text{d high} - \text{high} \times \text{d low} \) over \( \text{the square of what's below} \) is our derivative. So, let's write this out using our tool there.

So, \(\text{low} \times \text{d high} \) - that's my low function, \( 3 + x^2 \). \( \text{d high} \) is the derivative of that top function. I'm not going to take that derivative yet. I'm just going to write it out. \( \frac{d}{dx} \) of \( (2x - 1)^4 \).

So, that's my first term, minus high, my high function, \( (2x - 1)^4 \). \( \text{D low} \), that's the derivative of that bottom function, \( \frac{d}{dx} \) of \( 3 + x^2 \). Then, all of that gets divided by the square of what's below. Now, it's important here that you remember that this is already squared in our original function. But we need to square it again because it's \( \text{the square of what's below} \).

So we need to take that full term and square it as well. So, don't forget that this can be easy to mess up because you already have that squared there, but we have squared it. So let's go ahead and actually take our derivatives here because we've set up our quotient rule and now it remains to take these derivatives. Looking at this, I have 2 derivatives that I need to take, \( \frac{d}{dx} \) of \( (2x - 1)^4 \) and \( \frac{d}{dx} \) of \( 3 + x^2 \). For both of these terms, I'm going to have to use the chain rule.

Now remember that when doing these problems, it's going to be very important to stay organized. Make sure that you have enough space on your page to fully work it out and make sure that you are carrying down all the terms. So first, I have this \( 3 + x^2 \), and I don't need to do anything there. And then I'm going to multiply that by now taking the derivative of \( (2x - 1)^4 \). Now, using the chain rule here, I see that I have this inside function, \( 2x - 1 \).

And I have this outside function because all of that is raised to the power of 4. Now because that is raised to the power of 4, I'm going to start there with my outside function, starting from the outside, working our way in. So I'm going to go ahead and pull that 4 out to the front, decrease that power by 1 using my power rule. This becomes 4. I'm going to keep all of this in brackets just to keep it separated right now.

Four times all of that stuff on the inside, which is \( 2x - 1 \). Remember, when using the chain rule, we're treating that inside function like it's a variable until we get to our next step here. So \( 2x - 1 \), 4 times \( 2x - 1 \) to the power of 4 minus 1, which is 3. Now from here, we are not done using the chain rule. We are working our way inside.

Coming to that inside function, \( 2x - 1 \), the derivative of that inside function is just going to be 2. So, all of that is just multiplied by 2, and that is that first derivative there. But we are not done. We still have a lot more terms to go. So I'm subtracting here \( (2x - 1)^4 \), making sure to carry everything down that I need to.</

Intro to the Chain Rule Example 3

In this problem, we're asked to find the derivative of two different functions. Our first one is f(x) = sin⁵(x), and our second function is sin(x⁵). Now we want to find both of these derivatives, which we can do using the chain rule. So let's look at our first function here, sin⁵(x). Now, this might not be immediately clear how we're going to use the chain rule here, just because of how this is written.

But remember that if I have the function sin²(x), this is equivalent to saying sin(x), that whole function, squared. We just usually write it with that power, that exponent, in the middle of the sin and the x. So here, when I have sin⁵(x), I can also think of this as being sin(x) all to the power of 5. This makes it much more clear what our inside and outside functions are so that we can use the chain rule. So let's go ahead and get started here using our chain rule to find f'(x), our derivative.

Now here I have 2 functions. Of course, I have my inside function, which is sin(x), and I have my outside function, which is all of this stuff raised to the power of 5. Now, even though this has a trig function, our chain rule still works in the same exact way. We're going to start from the outside and work our way inside. So starting with that outside function here, we have all the stuff raised to the power of 5.

We're going to pull that 5 out to the front using the power rule and decrease that power by 1. So this becomes 5 times the sin(x), that's what my inside function is, all raised to the power of 4. Now working our way inside, we want actually to take the derivative of that inside function sin(x) and multiply by it.

Now, what is the derivative of sin(x)? We know that it is cos(x). So this is all multiplied by cos(x). Now we can rewrite this slightly because we know that the sin⁴(x), we're going to pull that 4 back inside just to write it by convention. So we have that 5 on the outside.

I'm actually going to write my cos(x) first here since it's not raised to a power. I have 5 cos(x) times sin⁴(x). And this is my final derivative, 5 cos(x) times sin⁴(x). Now that is our first function, we found the derivative here. Let's move on to our second function.

Now our second function here is still going to be using the chain rule. We have sin(x⁵). So here, it's a little bit more clear initially that I have my inside function x⁵, and I have my outside function, sin. So let's go ahead and see what this looks like when we apply the chain rule to find f'(x). Now, even though we have a trig function on the outside, we are still starting from the outside working our way in.

So we want to start by finding the derivative of that outside function. Now here, our outside function is sin. What is the derivative of sin? It is cos. So here, I have cos.

What am I taking the cos of? Well, since we're using the chain rule here, remember that we're keeping our inside function the same. So here, I'm taking the cos of x⁵, keeping that argument as is because that is my inside function. But we are not done yet because using the chain rule, we also need to take the derivative of that inside function and multiply by it. So I have cos(x⁵).

I am multiplying that by the derivative of x⁵. Using the power rule, moving that 5 to the front, decreasing that power by 1, that gives me 5x⁴. And this is my derivative. Now, conventionally, we typically want to write that variable and the constant upfront, 5x⁴ times cos(x⁵). So this is my final answer for my derivative here that I also found using the chain rule.

Now, it's going to be really important here to keep getting more practice because you want to get really, really comfortable using the chain rule no matter if you have trig functions or not. Let me know if you have any questions, and I will see you in the next one.

Find the derivative of the function.

$f\left(x\right)=\sin \text{⁡}\left(3x^2\right)$

• A. $\cos\left(6x\right)$
• B. $-6x\cos\left(3x^2\right)$
• C. $6x\cos \left(3x^2\right)$
• D. $\cos \left(3x^2\right)$

Find the derivative of the function.

$y=3{\cos }^4\text{⁡}\theta$

• A. $-4\sin^3\theta$
• B. $-12\sin^3\theta$
• C. $4\sin\theta\cos^3\theta$
• D. $-12\sin\theta\cos^3\theta$

Find the derivative of the function.

$f\left(t\right)=\sec (4t+5)$

• A. $4\sec\left(4t+5\right)$
• B. $4\sec\left(4t+5\right)\tan\left(4t+5\right)$
• C. $4\sec\left(t\right)\tan\left(t\right)$
• D. $\sec\left(4\right)\tan\left(4\right)$

The Chain Rule for 3+ Functions

We just learned that in order to find the derivative of a composite function, we can use the chain rule working our way from outside to inside. Now you may actually need to use the chain rule multiple times and this may sound like it's going to get a bit complicated, but we're going to tackle this in the same exact way, just working our way from outside to inside. So I'm going to show you exactly how this works here, and you'll be a pro in no time. So let's go ahead and jump right into our example here where we're asked to find the derivative of f(x)=sin4(3x2). It may not be immediately clear how we're going to use the chain rule multiple times, but we actually have 3 functions going on here.

We have 3x2. We have sine. And all of this is raised to the power of 4. Now we can actually rewrite this function in a way that will make it a bit easier to see how we're going to use the chain rule because whenever we have these trig functions that are raised to a power, like here, we have sine to the power of 4, we can rewrite our power on the outside of that function. So I can take my n value, whatever power raised to, and put it on the outside.

So with this function here, I can rewrite this as f(x)=(sin(3x2))4, just pulling that 4 to the outside. Now it's a bit more clear how we're going to use the chain rule because we can see we have this innermost function, this 3x2, and that is contained within this sine function, which is all inside of this raised to the 4th power. So we can see here that we can start from the outside and work our way inside. So let's go ahead and get started in finding this derivative, f′(x). Now we know we need to start with that outermost function, which is all of this stuff raised to the 4th power.

Now, it's really easy to get ahead of yourself here and start to worry about all of this stuff on the inside too. But remember that when we're working with that outside function, we're treating everything inside as though it's just a variable, as though all of this stuff is just x. So let's work with this outside function here, finding the derivative of something raised to the power of 4. We know that we can just use the power rule. So I'm going to pull that 4 out to the front there, multiplying by it and then decrease that power by 1.

So here this gives me 4 times all of this stuff that's on the inside, treating it as though it's just a variable right now, not worrying about any of those derivatives, all raised to the power of 4 minus 1 or 3. Now from here, we're continuing to work our way inside. And as we work our way inside, our next function here is the sine. So we want to multiply it by the derivative of that next function on the inside, sine. So multiplying here by the derivative of sine, we know that the derivative of sine is cosine.

So I'm multiplying this by the cosine, treating everything that's inside of that sine function, here, 3x2, as though it's just a variable. Now we have one more function here to worry about working our way inside to this 3x2. So, again, here, I'm going to multiply it by this derivative. Now the derivative of 3x2 is going to give me 6x, and this is my full derivative having used the chain rule. I started from the outside and then multiplied by each derivative working my way in.

Now we can do a little bit of simplification here because I have this 4 and this 6x that I can go ahead and multiply, giving me 24x. And then I'm also going to move this 3, this power back in just to clean up this notation a little bit. So this is multiplied by sin3(3x2) and then my final term here, that cos(3x2). And this is my full derivative here, f′(x), having used the chain rule multiple times. Now if you have any questions here, feel free to let me know, and let's continue getting practice with the chain rule.

I'll see you in the next video.

Find the derivative of the function.

$f(x)=\sin^5(2x^3+1)$

• A. $30x^2\cos⁡(2x^3+1)\sin^4⁡(2x^3+1)$
• B. $5⁡\sin^4(2x^3+1)$
• C. $5\cos^4(6x^2)$
• D. $30x^2\cos^4⁡\left(2x^3+1\right)$

Find the derivative of the function.

$y=\cos^3\left(\sec\theta\right)$

• A. $3\cos^2\left(\sec\theta\right)\sec\theta\tan\theta$
• B. $-3\sec \theta \tan \theta \cdot \sin (\sec \theta ){\cos }^2(\sec \theta )$
• C. $-3\sin^2\left(\sec\theta\tan\theta\right)$
• D. $-3\cos^2\left(\sec\theta\tan\theta\right)$

The Chain Rule for 3+ Functions Example 1

In this problem, we're asked to find the derivative of \( f(x) = \tan(3 - \sin(2x)) \). Now, this is a little different than some of the problems that we have been working with because we don't have anything raised to a power. However, we are still working with multiple functions here, and we are going to have to apply the chain rule multiple times. Let's go ahead and break this down. Looking at this function, my outermost function is the tangent.

Now working my way inside, I see that inside the argument of tangent, I have the function \( 3 - \sin(2x) \). So, what is my innermost function? Well, I see that inside the argument of sine, I have \( 2x \). So that is actually my innermost function. It might be a little clearer here if I add an extra set of parentheses around that \( 2x \) showing that this is my innermost function.

Let's go ahead and use our chain rule here to find our derivative \( f'(x) \). Remember that when using the chain rule, we are always starting from the outside and working our way in. Here, our outermost function is the tangent. So what is the derivative of tangent? The derivative of the tangent is \( \sec^2 \).

Since we're using the chain rule, whatever is inside the function remains unchanged. So my argument is still \( 3 - \sin(2x) \). This is where the argument for the first derivative ends. Working our way further inside, I consider the next function, \( 3 - \sin(2x) \).

The derivative of \( 3 \) is just \( 0 \), so I don’t have to worry about that. There is also a negative sign. The derivative of \( \sin(2x) \) is \( \cos(2x) \). Remember, whatever is in parentheses or within that function remains the same. One final step here, as we need to account for that innermost function \( 2x \) by taking its derivative and multiplying by it. The derivative of \( 2x \) is just \( 2 \). So, my last step here is to multiply by that final derivative of \( 2 \). Now we can simplify this a little bit and rewrite it to make it look a bit prettier.

Our derivative function, \( f'(x) \), pulling that constant up to the front and also moving this cosine to the front. The expression is \( -2 \cos(2x) \cdot \sec^2(3 - \sin(2x)) \), using the same argument that we started with here. If you wrote this in a different order, that's totally fine. It is still an equivalent statement.

This is our final derivative, and I will see you in the next video.