3: The Chain Rule

Example $\PageIndex{1}$: Using the Chain and Power Rules

Find the derivative of $h(x)=\dfrac{1}{(3x^2+1)^2}$.

Solution

First, rewrite $h(x)=\dfrac{1}{(3x^2+1)^2}=(3x^2+1)^{−2}$.

Applying the power rule with $g(x)=3x^2+1$, we have

$h'(x)=−2(3x^2+1)^{−3}\cdot 6x$.

Rewriting back to the original form gives us

$h'(x)=\dfrac{−12x}{(3x^2+1)^3}$

Example $\PageIndex{2}$: Using the Chain and Power Rules with a Trigonometric Function

Find the derivative of $h(x)=\sin^3x$.

Solution

First recall that $\sin^3x=(\sin x)^3$, so we can rewrite $h(x)=\sin^3x$ as $h(x)=(\sin x)^3$.

Applying the power rule with $g(x)=\sin x$, we obtain

$h'(x)=3(\sin x)^2\cos x=3\sin^2x\cos x$.

Example $\PageIndex{3}$: Finding the Equation of a Tangent Line

Find the equation of a line tangent to the graph of $h(x)=\dfrac{1}{(3x−5)^2}$ at $x=2$.

Solution

Because we are finding an equation of a line, we need a point. The $x$-coordinate of the point is 2. To find the $y$-coordinate, substitute 2 into $h(x)$. Since $h(2)=\dfrac{1}{(3(2)−5)^2}=1$, the point is $(2,1)$.

For the slope, we need $h'(2)$. To find $h'(x)$, first we rewrite $h(x)=(3x−5)^{−2}$ and apply the power rule to obtain

$h'(x)=−2(3x−5)^{−3}(3)=−6(3x−5)^{−3}$.

By substituting, we have $h'(2)=−6(3(2)−5)^{−3}=−6.$

Therefore, the line has equation $y−1=−6(x−2)$. Rewriting, the equation of the line is $y=−6x+13$.

Example $\PageIndex{4}$: Using the Chain Rule on a General Cosine Function

Find the derivative of $h(x)=\cos\big(g(x)\big).$

Solution

Think of $h(x)=\cos\big(g(x)\big)$ as $f\big(g(x)\big)$ where $f(x)=\cos x$. Since $f'(x)=−\sin x$, we have $f'\big(g(x)\big)=−\sin\big(g(x)\big)$. Then we do the following calculation.

$$\begin{align*} h'(x)&=f'\big(g(x)\big)\cdot g'(x) & & \text{Apply the chain rule.} \\[4pt] &=−\sin\big(g(x)\big)\cdot g'(x) & & \text{Substitute}\; f'\big(g(x)\big)=−\sin\big(g(x)\big).\end{align*} \nonumber$$

Thus, the derivative of $h(x)=\cos\big(g(x)\big)$ is given by $h'(x)=−\sin\big(g(x)\big)\cdot g'(x).$

Example $\PageIndex{5}$: Using the Chain Rule on a Cosine Function

Find the derivative of $h(x)=\cos(5x^2).$

Solution

Let $g(x)=5x^2$. Then $g'(x)=10x$. Using the result from the previous example,

$h'(x)=−\sin(5x^2)⋅10x=−10x\sin(5x^2)$

Example $\PageIndex{7}$: Combining the Chain Rule with the Product Rule

Find the derivative of $h(x)=(2x+1)^5(3x−2)^7$.

Solution

First apply the product rule, then apply the chain rule to each term of the product.

$\begin{align*} h'(x)&=\dfrac{d}{dx}\big((2x+1)^5\big)⋅(3x−2)^7+\dfrac{d}{dx}\big((3x−2)^7\big)⋅(2x+1)^5 & & \text{Apply the product rule.}\\[4pt] &=5(2x+1)^4⋅2⋅(3x−2)^7+7(3x−2)^6⋅3⋅(2x+1)^5 & & \text{Apply the chain rule.}\\[4pt] &=10(2x+1)^4(3x−2)^7+21(3x−2)^6(2x+1)^5 & & \text{Simplify.}\\[4pt] &=(2x+1)^4(3x−2)^6(10(3x−2)+21(2x+1)) & & \text{Factor out }(2x+1)^4(3x−2)^6\\[4pt] &=(2x+1)^4(3x−2)^6(72x+1) & & \text{Simplify.} \end{align*}$

Rule: Chain Rule for a Composition of Three Functions

For all values of $x$ for which the function is differentiable, if

$k(x)=h\Big(f\big(g(x)\big)\Big),$

then

$k'(x)=h'\Big(f\big(g(x)\big)\Big)\cdot f'\big(g(x)\big)\cdot g'(x).$

In other words, we are applying the chain rule twice.

Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, remember, we can always work from the outside in, taking one derivative at a time.

Example $\PageIndex{8}$: Differentiating a Composite of Three Functions

Find the derivative of $k(x)=\cos^4(7x^2+1).$

Solution

First, rewrite $k(x)$ as

$k(x)=\big(\cos(7x^2+1)\big)^4$.

Then apply the chain rule several times.

$\begin{align*} k'(x)&=4(\cos(7x^2+1))^3\cdot\dfrac{d}{dx}\big(\cos(7x^2+1)\big) & & \text{Apply the chain rule.}\\[4pt] &=4(\cos(7x^2+1))^3(−\sin(7x^2+1))\cdot\dfrac{d}{dx}\big(7x^2+1\big) & & \text{Apply the chain rule.}\\[4pt] &=4(\cos(7x^2+1))^3(−\sin(7x^2+1))(14x) & & \text{Apply the chain rule.}\\[4pt] &=−56x\sin(7x^2+1)\cos^3(7x^2+1) & & \text{Simplify} \end{align*}$

Example $\PageIndex{9}$: Using the Chain Rule in a Velocity Problem

A particle moves along a coordinate axis. Its position at time t is given by $s(t)=\sin(2t)+\cos(3t)$. What is the velocity of the particle at time $t=\dfrac{π}{6}$?

Solution

To find $v(t)$, the velocity of the particle at time $t$, we must differentiate $s(t)$. Thus,

$$v(t)=s'(t)=2\cos(2t)−3\sin(3t).\nonumber$$ □

Example $\PageIndex{10}$: Using the Chain Rule with Functional Values

Let $h(x)=f\big(g(x)\big).$ If $g(1)=4,g'(1)=3$, and $f'(4)=7$, find $h'(1).$

Solution

Use the chain rule, then substitute.

$$\begin{align*} h'(1)&=f'\big(g(1)\big)\cdot g'(1) & & \text{Apply the chain rule.} \\[4pt] &=f'(4)⋅3 & &\text{Substitute}\; g(1)=4 \;\text{and}\;g'(1)=3. \\[4pt] &=7⋅3 & &\text{Substitute}\; f'(4)=7. \\[4pt] &=21 & &\text{Simplify.} \end{align*} \nonumber$$

Example $\PageIndex{11}$: Taking a Derivative Using Leibniz’s Notation I

Find the derivative of $y=\left(\dfrac{x}{3x+2}\right)^5.$

Solution

First, let $u=\dfrac{x}{3x+2}$. Thus, $y=u^5$. Next, find $\dfrac{du}{dx}$ and $\dfrac{dy}{du}$. Using the quotient rule,

$\dfrac{du}{dx}=\dfrac{2}{(3x+2)^2}$

and

$\dfrac{dy}{du}=5u^4$.

Finally, we put it all together.

$$\begin{align*} \dfrac{dy}{dx}&=\dfrac{dy}{du}⋅\dfrac{du}{dx} & & \text{Apply the chain rule.}\\[4pt] &=5u^4⋅\dfrac{2}{(3x+2)^2} & & \text{Substitute}\; \frac{dy}{du}=5u^4\;\text{and}\;\frac{du}{dx}=\frac{2}{(3x+2)^2}. \\[4pt] &=5\left(\dfrac{x}{3x+2}\right)^4⋅\dfrac{2}{(3x+2)^2} & & \text{Substitute}\; u=\frac{x}{3x+2}. \\[4pt] &=\dfrac{10x^4}{(3x+2)^6} & & \text{Simplify.} \end{align*}$$

It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.

Example $\PageIndex{12}$: Taking a Derivative Using Leibniz’s Notation II

Find the derivative of $y=\tan(4x^2−3x+1).$

Solution

First, let $u=4x^2−3x+1.$ Then $y=\tan u$. Next, find $\dfrac{du}{dx}$ and $\dfrac{dy}{du}$:

$\dfrac{du}{dx}=8x−3$ and $\dfrac{dy}{du}=\text{sec}^2u.$

Finally, we put it all together.

$$\begin{align*} \dfrac{dy}{dx}&=\dfrac{dy}{du}⋅\dfrac{du}{dx} & & \text{Apply the chain rule.} \\[4pt] &=\text{sec}^2u⋅(8x−3) & & \text{Use}\;\dfrac{du}{dx}=8x−3\;\text{and}\;\dfrac{dy}{du}=\text{sec}^2u. \\[4pt] &=\text{sec}^2(4x^2−3x+1)⋅(8x−3) & & \text{Substitute}\;u=4x^2−3x+1. \end{align*} \nonumber$$