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 plus 5. We also know how to find the derivative of some variable like x cubed. But what if we put those two together and we're asked to find the derivative of (4x + 5) cubed? How could we go about that? Well, we could go ahead and multiply out that 4x + 5 three 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 ddxfgx, 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, ddx4x+53. 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 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 times (4x + 5) squared. 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 Leibnitz 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 dydx, that's the derivative of my whole function, is equal to dydududx. 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, \( f(x) \) is equal to 2 times (3x squared - x) to the power of 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 squared - 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 times that inside function, remember, we're treating that like it's a variable, leaving it completely alone here, to the power of 3. Then from here, we want to multiply by the derivative of that inside function. Now here, the derivative of 3x squared - 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. $\frac{10x-3}{2\sqrt{5x^2-3}}$
• D. $\frac{1}{2\sqrt{5x^2-3}}$

Find the derivative of the function.

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

• A. $\frac32\sqrt{24x^2-2}$
• B. $\left(36x^2-3\right)\sqrt{8x^3-2x}$
• C. $\frac{3}{2\sqrt{24x^2-2}}$
• D. $\left(12x^2-1\right)\sqrt{8x^3-2x}$

Find the derivative of the function.

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

• 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 \( \frac{dy}{dx} \), our derivative for the function \( y = (2x - 1)^4 (3 + x^2) \). 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 multiplying 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. Now, this can get a little bit messy, so it's important here to stay organized and know what step you're on and not leave out any terms.

So because we have functions inside of 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 \( \frac{dy}{dx} \). So let's start here by first looking at these terms altogether. Because we have to use the product rule, these are being multiplied.

So what does that tell us here? Well, remember that when working with our product rule, we're going to take that left function, leave it as is. So I have \( (2x - 1)^4 \). Then we're gonna multiply that by the derivative of that function on the right. So that's a \(\frac{d}{dx}\) of \( 3+x^2 \).

Now I haven't actually applied my chain rule yet, but when we take this derivative, we are going to be using the chain rule. Now I'm going to add this together, still using my product rule here, times that right side function as is \( 3+x^2 \), keeping that as is, again, multiplying by the derivative of our other function, this time our left function. Remember, our product rule is always 1 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.

So now let's actually take these derivatives using the chain rule. We have set up our product rule because the big picture of this function is 2 functions being multiplied together. Now diving into each of these individual terms, we wanna actually take these derivatives. So, again, here we wanna 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 here, let's start using the chain rule. Remember here to use the chain rule, we want to look at what our inside function and what our outside function is. Here, our inside function we have is \( 3+x \). And then our outside function is all of that stuff being squared. So multiplying here.

I'm gonna go ahead and put a bracket because I know that there's gonna be a little bit more stuff right here when we take this derivative because we're using the chain rule. Now starting with that outside function, remember with the chain rule, no matter if we're using it in conjunction with the product rule or not, we're using it the same exact way. So here, starting with our outside function here and then working our way in, using the power rule for that outside function, we have \( 2 \times (3+x) \). All of that is raised to the power of \( 2-1 \) because we use the product rule here. \( 2-1 \) is just 1, so I don't really need to write that one there.

I can leave it as is. 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 a 1. So this is really just all being multiplied by 1, which ultimately won't do anything and I can eventually drop off that 1. But we want to stay organized and I'm going to write everything that I 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 \( 3+x^2 \). 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 4th power. So we're gonna pull that 4 out to the front using the power rule and then subtract 1. So this becomes \( 4 \times \text{that stuff in parentheses} \). 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 are 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, and we're all done here. Now this is technically a correct answer, but we're gonna need to simplify a little bit more because there is quite a bit of stuff going on.

So let's go ahead and do some simplification here. So I'm gonna keep that \( (2x - 1)^4 \) out front as is, and then I'm gonna do a little bit of combining here because I have 2 and I have 1. \( 2 \times 1 \) is just 2. So this is multiplying \( 2 \times (3+x) \). Now from here, I'm adding that together with \( 3+x^2 \), no simplification to be done to that term.

And then here I have a 4 and a 2. Multiplying that together gives me 8. This is \( 8 \times (2x - 1)^3 \). Now, again, this is still technically a correct answer, and how you simplify is largely dependent on what your instructor expects of you when simplifying. Remember that whenever you're solving these problems, if you don't get an answer that looks exactly like mine, it still may be an equivalent expression.

So always double check for different ways that we might have simplified stuff. So we can do a little bit more here because we can do some moving around just to make this look a little bit prettier. Typically, we want our constants out front of a term, so I'm going to go ahead and pull that constant out front. I'm also going to put that \( 3+x \) there and then multiply it by \( (2x - 1)^4 \), just because for convention, we want to move that around. And then here again, moving that 8 out to the front, I have \( 8 \times (3+x)(2x-1)^3 \).

Now, this is kind of a complicated derivative here, but this is our final answer. We have a lot going on here because we had to use a bunch of different rules at the same time. Now, we could do some factoring here, but it's really not going to make this look any nicer because we're still going to have the same number of terms. You could do a bit more factoring. Feel free to try that on your own.

But for now, this is our final derivative, \( \frac{dy}{dx} \), of our original function. Let me know if you have any questions here, and feel free to go ahead and erase all of your work, retry this on your own, and double-check it with me. Thanks for watching. I'll see you in the next one.

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 bit intimidating, and 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 that 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 go ahead and 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 also need to use the chain rule. So let's go ahead and break this down. Now in finding my derivative \( dy/dx \), we know we need to use the quotient rule, and the quotient rule tells us \( \frac{{low \cdot d(high) - high \cdot d(low)}}{{(low)^2}} \) is our derivative. So let's go ahead and write this out using our little tool there.

So \( low \) \( d(high) \) that's my low function, \( 3+x^2 \). \( d(high) \) 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, \( (2x - 1)^4 \). \( d(low) \), 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 the square of what's below.

So we need to take that full term and square it as well. So don't forget that 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 essential 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 of the terms. So first, I have this \( 3+x^2 \). I don't need to do anything there. And then I'm going to multiply that by now taking the derivative of this \( 2x-1 \) to the power of 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 \( 4 - 1 \), 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.

And now here, we're going to take another derivative using the chain rule. Now here, my inside function is \( 3+x \). My outside function is all of that stuff squared. So I'm going to go ahead and pull that 2 out to the front using the power rule, decrease that power by 1. That's going to give me 2 times, that stuff in the middle, \( 3+x \).

Now that would just be the power of 1, so I don't need to write that because it already is raised to the power of 1. Now from here, we are, of course, not done. We know that by now. \( 3+x \). We're going to take the derivative of that inside function.

We are working our way inside. The derivative of \( 3+x \) because I have a constant and I have \( x \), it's just going to be 1. So that is our full derivative here. Now remember not to drop your denominator. We still have a denominator here.

And we can go ahead and do a little bit of simplification because we know that if we take something squared, we raise it again to the power of 2. It is now going to be to the power of 4. So this is \( (3+x)^4 \). Now we can do some simplification. We've taken all the derivatives that we need to.

We know that sometimes the most annoying part of the quotient rule is doing all of that algebraic simplification. But what can we do here? So looking at everything that's going on, I know that I can go ahead and take care of these constants and multiply them out, pull them out to the front. And the other thing that I do see happening here is that I have some similar terms. So I have \( 3+x^2 \), and I have \( 3+x \) right here.

And on the bottom, I have \( 3+x \) raised to the power of 4, which tells me that I can do a little bit of canceling here to make this a little bit cleaner. So what all can we cancel? Well, since here I have this to the lowest power, that's all that we can cancel. We can cancel one \( 3+x \). So that means that if I cancel 1 of these \( 3+x's \) out of this term, this 2 is just going to become a 1, which is not something that we're going to have to worry about writing.

And then this 4, since I can only cancel one of them, is going to become a 3. So let's go ahead and do some rewriting here. Since I have this \( 3+x \) to the power of 1, that's just \( 3+x \), that is multiplying this 2 times 4. Remember, we are going to go ahead and multiply those constants together. That is 8 times \( 2x - 1 \) cubed.

Double-check that we have accounted for everything going on in this term. Now I'm going to subtract here \( (2x - 1)^4 \) times 2. Because, remember, this term fully canceled out. So this looks a little bit nicer here. Then all of that is being divided by \( (3+x)^3 \).

Now from here, the only thing that I want to do here is pull those constants out to the front. You could do a little bit more factoring here because \( 2x - 1 \) are common terms happening here. But factoring that out won't simplify this much because we're still going to have the same number of terms. If factoring is something that your instructor wants you to do every single time, no matter if it really makes it a simpler term or not, that's fine. You can go ahead and do that as well.

But all I'm going to do here is pull out my constants to the front to organize this a little bit better. So this becomes 8 times \( 3+x \) times \( 2x - 1 \) cubed minus, pulling that 2 out to the front, 2 times \( 2x - 1 \) to the power of 4. So it looks a little bit nicer here. And then I have that \( (3+x)^3 \) on the bottom, and this is my final answer for my derivative. It is a rather long and tedious process when you have to use the quotient rule already.

But now combining here the quotient rule with the chain rule, it can get a little bit messy. Remember to take your time on these problems and double-check that you are accounting for every single term, every single derivative. Thanks for watching. Let me know if you have questions.