Chain Rule

Suppose you were asked to differentiate the function $$f(x) = \sqrt{x^2 + 1}.$$ The differentiation formulas such as the Product Rule and Quotient Rule are not sufficient to evaluate this derivative, as this function cannot easily be expressed as a product or quotient of two functions.

However, unlike regular numbers, functions have one additional property -- that of composition. We can express this function $f(x)$ as a composition of two functions: $g(u) = \sqrt{u}$ and $h(x) = x^2 + 1$.

It in fact turns out that the derivative $f'(x)$ is just equal to the product of the derivatives of $g$ and $h$.

Thus:

This is in fact one of the most important of the differentiation formulas, and is called the Chain Rule. A somewhat vague proof involves interpreting derivatives as rates of change. Take $da/db$ to be the rate of change of $a$ with respect to $b$. For example, if $u$ changes twice as fast as $x$ changes and $y$ changes three times as fast as $u$ changes, we expect that $y$ changes six times as fast as $x$ changes, so that $$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}.$$

A more formal proof of the Chain Rule, as well as alternative explanations, can be found here. Now, on with the show!

Let's try to figure out the derivative of $(x+2)^2$. This is actually the composition of two functions! Let: $$f(x) = x+2$$ and $$g(x) = x^2.$$

The Chain Rule now states that if a function can be expressed as a composition like $g(f(x))$, then its derivative is exactly $$g'(f(x)) f'(x).$$

That's the chain rule!

Chain Rule from Bicycles

Let's start with something straightforward, which doesn't even involve Calculus!

(Don't worry, Calculus will materialize in due course.)

Let's do a quick calculation. See the feet on the pedals in the picture? The pedals are at some radius from the center of the big wheel.

Let's pretend it was exactly 4.

But what about people who are afraid of heights? Why not make the big wheel only half as tall? If we did that, and didn't change the pedal radius of the cycle, then...

That's a lot slower! You'd need to pedal twice as fast to keep up with the big-wheel guys, and you'd soon become The Road Runner.

Chains to the rescue

Today, bicycles have chains, which look something like this:

Now suppose that the radius of the small gear is $1/4$ the radius of the large (pedal) gear.

Now, let's add the wheel! It's attached to the small gear, so that whenever the small gear turns one full revolution, so does the wheel.

Indeed, when the pedal moves through distance $x$, the edge of the wheel moves through distance $h(f(x)) = 5x$. Let's define this to be $w(x) = 5x$.

Speed

Recall from above that the function $w(x)$ calculates the distance the edge of the wheel moves when the pedal moves through $x$.

Indeed, we've calculated that the edge of the wheel moves distance $5t$ in time $t$.

Now let $y(t) = w(p(t))$.

Since we calculated the speed to be 5 m/s above, we have found that $$y'(t) = 5.$$

Let's dig a bit deeper.

What, then, if the pedal distance function is something like $p(t) = t^2$?

Yes, we've used the fact that the gears, chain, and wheel are a machine that multiplies pedal speed by 5.

What if $p(t)$ were any other function? What's the general formula?

Recall that $w(x)$ computes how far the wheel edge moves when the pedal moves by $x$, and the total distance the wheel edge moves in time $t$ is $y(t) = w(p(t))$.

Hmm...

But...

Chain Rule

(This post is unfinished, but it can be finished by anyone in the community! In fact, you don't even need to finish the whole post - if you have an idea of how to write part of it, please take the baton and contribute!)

1. State the chain rule carefully as a theorem.

2. Explain that there's an issue when the gear mechanism doesn't simply multiply speed by 5 no matter what time it is. Example: a dented wheel (rest of chain system intact).

3. With the dented wheel, suppose that at some area of the wheel, the radius is less than 10 times that of the small gear. Note that when the wheel is still on the undented part, the chain machine instantaneously does multiply the speed by 5.

4. That motivates why we should multiply derivatives of w and p.

5. Ask why we need to evaluate the derivative of w at p(t), and explain that part.

That's the chain rule! :-)

General formula for functions $f(x)$ and $g(x)$:

$\frac{d}{dx}(f(g(x)) = f'(g(x))g' (x)$

Solution: Since the derivative of the outside function is $\cos 2x$ and the derivative of the inside function is $2$:

$\frac{d}{dx}(\sin(2x)) = 2\cos(2x)$

See Pete Clark's physical intuition here, at point (2).

For a proof using linear approximation and (implicitly) little-o notation, see this proof on Wikipedia. (In fact, see if you can understand the slopes or "sensitivities" of the linear approximations in the proof through the aforementioned physical intuition of "multiplying rates".)

The formula for the Chain Rule for single-variable functions is $f(g(x))=f'(g(x))g'(x)$. If we have an equation $f(g(x))=(x+2)^3$, we first determine the "outside" part of the function and the "inside" part of the function. The "outside" part of the function is $x^3$ and the "inside" part of the function is $x+2$, therefore, by the formula, $f(x)=g(x)^3$ and $g(x)=x+2$.

Next, we take the derivative of the outside while keeping the inside the same, and then we take the derivative of the inside function and multiply them together, so it should look like this: $f'(g(x))g'(x)=3(x+2)^2 (1)$.

In this case, the derivative of the inside function is 1 so if we simplify it, we can just take it out (because anything multiplied by 1 doesn't change it). Simplified, it would just be $f'(g(x))g'(x)=3(x+2)^2$.

The Chain Rule for multi-variable functions is a bit more complicated. It involves partial derivatives.

We must first take the partial derivative with respect to $x$, multiply it by the derivative of the inside function with respect to $t$, then add it to the partial derivative with respect to $y$ and multiply that by the inside function with respect to $t$.

After that, we substitute all of the $x$'s and $y$'s for $t$'s so that the whole answer is in terms of $t$. An example: $f(x,y)=3x^2+4y$ $x=\sin(t)$ $y=3t$ The partial derivative of $f(x,y)$ with respect to $x$ is $6x$. The derivative of $x$ with respect to $t$ is $\cos(t)$. The partial derivative of $f(x,y)$ with respect to $y$ is 4. The derivative of $y$ with respect to $t$ is 3. We then start to put it together: $f'(x,y)=6x(\cos(t))+4(3)$ Since it is not entirely in terms of $t$ (there is an $x$ term still in there) we put $\sin(t)$ in the place of $x$ because $x=\sin(t)$. Now we have the final answer: $f'(x,y)=6\sin(t)(\cos(t))+12$

Suppose we need to take the derivative of $\sqrt{8x^2+10}$. We can do this by using the chain rule. In order to complete the chain rule it is always easier to rewrite the function in exponential form. For example, $\sqrt{x}$ would be rewritten as $x^{1/2}$.

First, rewrite $\sqrt{8x^2+10}$ in exponential form.

$(8x^2+10)^{1/2}$

The next step is to execute the chain rule. I like to think of this rule as an "inside and out" method. What we will do now is** take the derivative of the "outside"** of the function then take the derivative of the "inside" of the function and multiply them together. For this it helps to think of the outside of the function as $z^{1/2}$ where $z=8x^2+10$.

First, take the derivative of the outside.

$\frac{1}{2}\,z^{-1/2}$

Now take the derivative of the inside.

$16x$

Now multiple the outside and the inside together.

$8x(z^{-1/2})$

And substitute in for $z$.

$8x(8x^2+10)^{-1/2}$ $=$ $8x/\sqrt{8x^2+10}$

To recap

1. Rewrite the function in exponential form
2. Take the derivative of the "outside" of the function
3. Take the derivative of the "inside" of the function
4. Multiply the derivative of the outside of the function by the derivative of the inside of the function

Here is a worksheet containing chain rule problems. Answers are at the bottom. Good luck!

Derivative of Inverse

If you got the answer, great job. Let's walk through the solution.

The chain rule allows us to find the derivatives of compositions of functions.

Indeed, let $t = x^2+1$, and $f(x) = \frac{1000}{t}$. Now we can apply the chain rule.

What is $\frac{d}{dt} \frac{1000}{t}$?

The chain rule says that the derivative of $f$ is the product of these 2 intermediate computations.

Great work. Now, substituting $x = 3$, we find that the slope is $-60$.

(Videos) Chain Rule Problems

by PatrickJMT

The Chain Rule is a very simple and important rule in derivatives that can get very complicated and messy very quickly. Below are a bunch of videos by patrickJMT showing you different examples of the chain rule.

Here's a combination of the product rule and the chain rule.

Here's a combination of the quotient rule and the chain rule.

Here are some more complex examples of the chain rule.

(Videos) Using Chain, Product, and Quotient Rules to Find Derivatives

by PatrickJMT

Here's a video by patrickJMT showing you how to use the chain, product, and quotient rules to find the derivatives of functions.

Recall the chain, product, and quotient rules: $$\begin{align} [f(g(x))]'&=f'(g(x))\cdot g'(x)\\ [f(x)g(x)]'&=f'(x)g(x)+g'(x)f(x)\\ \bigg[\frac{f(x)}{g(x)}\bigg]'&=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2} \end{align}$$

The first problem he finds the derivative of is: $$f(x)=(2x^{\frac13}-6x^{\frac53})^3\cdot(4x^2-5x)^{\frac13}$$

I like to calculate each pieces derivative separately and then plug them into the formula after. It helps me keep things organized.

So I'll start by finding the derivative of the first part. $$f_1(x)=(2x^{\frac13}-6x^{\frac53})^3$$ Here's where we use the chain rule. Take the derivative of the outside and then multiply it by the derivative of the inside. $$\begin{align} f_1'(x)&=\color{red}{3}(2x^{\frac13}-6x^{\frac53})^{\color{red}{3-1}}\color{red}{\cdot(2x^{\frac13}-6x^{\frac53})'}\\ &=3(2x^{\frac13}-6x^{\frac53})^2\cdot[2(\color{red}{\frac13}x^{\color{red}{\frac13-1}})-6(\color{red}{\frac53}x^{\color{red}{\frac53-1}})]\\ &=\boxed{3(2x^{\frac13}-6x^{\frac53})^2\cdot\bigg(\frac23x^{-\frac23}10x^{\frac23}\bigg)} \end{align}$$

Now let's find the derivative of the second part. $$f_2(x)=(4x^2-5x)^{\frac13}$$ Again, we need to use the chain rule. $$\begin{align} f_2'(x)&=\color{red}{\frac13}(4x^2-5x)^{\color{red}{\frac13-1}}\color{red}{\cdot(4x^2-5x)'}\\ &=\frac13(4x^2-5x)^{-\frac23}\cdot[4(\color{red}{2}x^{\color{red}{2-1}})-5(\color{red}{1}x^{\color{red}{1-1}})]\\ &=\boxed{\frac13(4x^2-5x)^{-\frac23}\cdot(8x-5)} \end{align}$$

Now that we have all the pieces we can plug them into our product rule for derivatives formula. $$\begin{align} f'(x)&=f_1'(x)\cdot f_2(x)+f_2'(x)\cdot f_1(x)\\ &=3(2x^{\frac13}-6x^{\frac53})^2\cdot\bigg(\frac23x^{-\frac23}10x^{\frac23}\bigg)\cdot(4x^2-5x)^{\frac13}+\\ &\ \ \ \ \ +\frac13(4x^2-5x)^{-\frac23}\cdot(8x-5)\cdot(2x^{\frac13}-6x^{\frac53})^3\\ &=3(2x^{\frac13}-6x^{\frac53})^2\cdot\bigg(\frac23x^{-\frac23}10x^{\frac23}\bigg)\cdot(4x^2-5x)^{\frac13}+\\ &\ \ \ \ \ +\frac{(8x-5)(2x^{\frac13}-6x^{\frac53})^3}{3(4x^2-5x)^{\frac23}} \end{align}$$

Here's another example of solving a derivative using the chain and quotient rules.

The problem he looks at is: $$g(x)=\frac{(2x^3-5x)^4}{(4x+1)^5}$$

I like to calculate each pieces derivative separately and then plug them into the formula after. It helps me keep things organized.

Let's start with the top piece's derivative. $$g_1(x)=(2x^3-5x)^4$$ Here we have to take the derivative outside the parentheses and then multiply on the derivative of inside the parentheses. $$\begin{align} g_1'(x)&=\color{red}{4}(2x^3-5x)^{\color{red}{4-1}}\color{red}{\cdot(2x^3-5x)'}\\ &=4(2x^3-5x)^3\cdot(2(\color{red}{3}x^{\color{red}{3-1}})-5(\color{red}{1}x^{\color{red}{1-1}}))\\ &=\boxed{4(2x^3-5x)^3\cdot(6x^2-5)} \end{align}$$

Now let's derive the bottom piece. $$g_2(x)=(4x+1)^5$$ Again, we use the chain rule to find this derivative. $$\begin{align} g_2'(x)&=\color{red}{5}(4x+1)^{\color{red}{5-1}}\color{red}{\cdot(4x+1)}\\ &=5(4x+1)^4\cdot(4(\color{red}{1}x^{\color{red}{1-1}})+\color{red}{0})\\ &=5(4x+1)^4\cdot(4)\\ &=\boxed{20(4x+1)^4} \end{align}$$

So now we can plug these into our quotient rule derivative formula. $$\begin{align} g'(x)&=\frac{g_1'(x)\cdot g_2(x)-g_2'(x)\cdot g_1(x)}{(g_2(x))^2}\\ &=\frac{4(2x^3-5x)^3\cdot(6x^2-5)\cdot(4x+1)^5-20(4x+1)^4\cdot(2x^3-5x)^4}{[(4x+1)^5]^2}\\ &=\frac{4(2x^3-5x)^3\cdot(6x^2-5)\cdot(4x+1)^5-20(4x+1)^4\cdot(2x^3-5x)^4}{(4x+1)^{10}}\\ &=\frac{4(4x+1)^4(2x^3-5x)^3[(6x^2-5)\cdot(4x+1)-5(2x^3-5x)]}{(4x+1)^{10}}\\ &=\frac{4(2x^3-5x)^3[(6x^2-5)\cdot(4x+1)-5(2x^3-5x)]}{(4x+1)^{6}}\\ &=\frac{4(2x^3-5x)^3[24x^3-20x+6x^2-5-10x^3+25x]}{(4x+1)^{6}}\\ &=\boxed{\frac{4(2x^3-5x)^3[14x^3+6x^2+5x-5]}{(4x+1)^{6}}} \end{align}$$

Here's a third video where he works through another derivative problem involving chain rule.

The problem he looks at this time is: $$h(x)=\frac{(x+1)^2}{x}-\bigg(\frac{2}{x^3+1}\bigg)^2-5x$$ He first simplifies this problem before taking the derivative in order to make it easier. $$\begin{align} &=\frac{x^2+2x+1}{x}-\frac{4}{(x^3+1)^2}-5x\\ &=x+2+x^{-1}-4(x^3+1)^{-2}-5x \end{align}$$ Now we can take the derivative. $$\begin{align} h'(x)&=1+0+(-1x^{-2})-4[-2(x^3+1)^{-3}]\cdot(3x^2)-5\\ &=1-x^{-2}+8(x^3+1)^{-3}(3x^2)-5\\ &=\boxed{-4-\frac{1}{x^2}+\frac{24x^2}{(x^3+1)^3}} \end{align}$$

(Video) Chain Rule for Finding Derivatives

by PatrickJMT

Here's a video by patrickJMT explaining the chain rule and how to use it to find derivatives.

The chain rule can be though of as the outside-inside rule. You take the derivative of a function that's actually a composition of functions. It's the derivative of $f$ where the inside is left alone, multiplied by the derivative of the inside. $$h(x)=f[g(x)]\\h'(x)=f'[g(x)]\cdot g'(x)$$

The first problem that he looks at is: $$y=\sin x^2$$ Here the outside function is the sine function. The inside function is $x^2$. So we take the derivative of the outside (leaving the inside alone) and then multiply on the derivative of the inside. $$\begin{align} y'&=\cos x^2\cdot\color{red}{(x^2)'}\\ &=\boxed{\cos x^2\cdot2x} \end{align}$$

The second problem is: $$y=(x^2+3x+4)^{30}$$ So here the outside function is the parentheses being raised to the $30$-th power. The inside function is $x^2+3x+4$. $$\begin{align} y'&=30(x^2+3x+4)^{29}\cdot\color{red}{(x^2+3x+4)'}\\ &=\boxed{30(x^2+3x+4)^{29}\cdot(2x+3)} \end{align}$$

(Videos) Chain Rule Examples

by PatrickJMT

Here's a video by patrickJMT where he works through some examples involving using the chain rule to find derivatives.

Recall: $$y=f[g(x)]\\ y'=f'[g(x)]\cdot g'(x)$$

The first example that he solves is: $$f(x)=6(x^3+2x)^{100}$$ $$\begin{align} f'(x)&=6*100(x^3+2x)^{99}\cdot\color{red}{(x^3+2x)'}\\ &=600(x^3+2x)^{99}\cdot(3x^2+2)\\ &=\boxed{600(3x^2+2)(x^3+2x)^{99}} \end{align}$$

In this video he looks at another example. $$\begin{align} g(x)&=\frac{(x^2-1)^4}{2}-6(x+7)^4\\ &=\frac12(x^2-1)^4-6(x+7)^4\\ g'(x)&=\frac12[4(x^2-1)^3\color{red}{(x^2-1)'}]-6[4(x+7)^3\color{red}{(x+7)'}]\\ &=2(x^2-1)^3(2x)-24(x+7)^3(1)\\ &=\boxed{4x(x^2-1)^3-24(x+7)^3} \end{align}$$

The third problem he looks at has a nested chain rule; a chain rule within a chain rule. Don't fret. Just take it slow and be careful. $$h(x)=[5x-(1+x^2)^3]^9$$ Again, we work from the outside in. $$\begin{align} h'(x)&=9[5x-(1+x^2)^3]^8\color{red}{[5x-(1+x^2)^3]'}\\ &=9[5x-(1+x^2)^3]^8\color{red}{[(5x)'-((1+x^2)^3)']} \end{align}$$ Now we come to the second chain rule. $$\begin{align} h'(x)&=9[5x-(1+x^2)^3]^8[5-3(1+x^2)^2\color{red}{(1+x^2)'}]\\ &=9[5x-(1+x^2)^3]^8[5-3(1+x^2)^2(2x)]\\ &=\boxed{9[5x-(1+x^2)^3]^8[5-6x(1+x^2)^2]} \end{align}$$