# 3.6 The chain rule  (Page 5/6)

 Page 5 / 6

## Taking a derivative using leibniz’s notation, example 1

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

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

$\frac{du}{dx}=\frac{2}{{\left(3x+2\right)}^{2}}$

and

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

Finally, we put it all together.

$\begin{array}{ccccc}\hfill \frac{dy}{dx}& =\frac{dy}{du}·\frac{du}{dx}\hfill & & & \text{Apply the chain rule.}\hfill \\ & =5{u}^{4}·\frac{2}{{\left(3x+2\right)}^{2}}\hfill & & & \text{Substitute}\phantom{\rule{0.2em}{0ex}}\frac{dy}{du}=5{u}^{4}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{du}{dx}=\frac{2}{{\left(3x+2\right)}^{2}}.\hfill \\ & =5{\left(\frac{x}{3x+2}\right)}^{4}·\frac{2}{{\left(3x+2\right)}^{2}}\hfill & & & \text{Substitute}\phantom{\rule{0.2em}{0ex}}u=\frac{x}{3x+2}.\hfill \\ & =\frac{10{x}^{4}}{{\left(3x+2\right)}^{6}}\hfill & & & \text{Simplify.}\hfill \end{array}$

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.

## Taking a derivative using leibniz’s notation, example 2

Find the derivative of $y=\text{tan}\phantom{\rule{0.1em}{0ex}}\left(4{x}^{2}-3x+1\right).$

First, let $u=4{x}^{2}-3x+1.$ Then $y=\text{tan}\phantom{\rule{0.1em}{0ex}}u.$ Next, find $\frac{du}{dx}$ and $\frac{dy}{du}\text{:}$

$\frac{du}{dx}=8x-3\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{dy}{du}={\text{sec}}^{2}u.$

Finally, we put it all together.

$\begin{array}{ccccc}\hfill \frac{dy}{dx}& =\frac{dy}{du}·\frac{du}{dx}\hfill & & & \text{Apply the chain rule.}\hfill \\ & ={\text{sec}}^{2}u·\left(8x-3\right)\hfill & & & \text{Use}\phantom{\rule{0.2em}{0ex}}\frac{du}{dx}=8x-3\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{dy}{du}={\text{sec}}^{2}u.\hfill \\ & ={\text{sec}}^{2}\left(4{x}^{2}-3x+1\right)·\left(8x-3\right)\hfill & & & \text{Substitute}\phantom{\rule{0.2em}{0ex}}u=4{x}^{2}-3x+1.\hfill \end{array}$

Use Leibniz’s notation to find the derivative of $y=\text{cos}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right).$ Make sure that the final answer is expressed entirely in terms of the variable $x.$

$\frac{dy}{dx}=-3{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right)$

## Key concepts

• The chain rule allows us to differentiate compositions of two or more functions. It states that for $h\left(x\right)=f\left(g\left(x\right)\right),$
${h}^{\prime }\left(x\right)={f}^{\prime }\left(g\left(x\right)\right){g}^{\prime }\left(x\right).$

In Leibniz’s notation this rule takes the form
$\frac{dy}{dx}=\frac{dy}{du}·\frac{du}{dx}.$
• We can use the chain rule with other rules that we have learned, and we can derive formulas for some of them.
• The chain rule combines with the power rule to form a new rule:
$\text{If}\phantom{\rule{0.2em}{0ex}}h\left(x\right)={\left(g\left(x\right)\right)}^{n},\text{then}\phantom{\rule{0.2em}{0ex}}{h}^{\prime }\left(x\right)=n{\left(g\left(x\right)\right)}^{n-1}{g}^{\prime }\left(x\right).$
• When applied to the composition of three functions, the chain rule can be expressed as follows: If $h\left(x\right)=f\left(g\left(k\left(x\right)\right)\right),$ then ${h}^{\prime }\left(x\right)={f}^{\prime }\left(g\left(k\left(x\right)\right){g}^{\prime }\left(k\left(x\right)\right){k}^{\prime }\left(x\right).$

## Key equations

• The chain rule
${h}^{\prime }\left(x\right)={f}^{\prime }\left(g\left(x\right)\right){g}^{\prime }\left(x\right)$
• The power rule for functions
${h}^{\prime }\left(x\right)=n{\left(g\left(x\right)\right)}^{n-1}{g}^{\prime }\left(x\right)$

For the following exercises, given $y=f\left(u\right)$ and $u=g\left(x\right),$ find $\frac{dy}{dx}$ by using Leibniz’s notation for the chain rule: $\frac{dy}{dx}=\frac{dy}{du}\phantom{\rule{0.2em}{0ex}}\frac{du}{dx}.$

$y=3u-6,u=2{x}^{2}$

$y=6{u}^{3},u=7x-4$

$18{u}^{2}·7=18{\left(7x-4\right)}^{2}·7$

$y=\text{sin}\phantom{\rule{0.1em}{0ex}}u,u=5x-1$

$y=\text{cos}\phantom{\rule{0.1em}{0ex}}u,u=\frac{\text{−}x}{8}$

$\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}u·\frac{-1}{8}=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\frac{\text{−}x}{8}\right)·\frac{-1}{8}$

$y=\text{tan}\phantom{\rule{0.1em}{0ex}}u,u=9x+2$

$y=\sqrt{4u+3},u={x}^{2}-6x$

$\frac{8x-24}{2\sqrt{4u+3}}=\frac{4x-12}{\sqrt{4{x}^{2}-24x+3}}$

For each of the following exercises,

1. decompose each function in the form $y=f\left(u\right)$ and $u=g\left(x\right),$ and
2. find $\frac{dy}{dx}$ as a function of $x.$

$y={\left(3x-2\right)}^{6}$

$y={\left(3{x}^{2}+1\right)}^{3}$

a. $u=3{x}^{2}+1;$ b. $18x{\left(3{x}^{2}+1\right)}^{2}$

$y={\text{sin}}^{5}\left(x\right)$

$y={\left(\frac{x}{7}+\frac{7}{x}\right)}^{7}$

a. $f\left(u\right)={u}^{7},u=\frac{x}{7}+\frac{7}{x};$ b. $7{\left(\frac{x}{7}+\frac{7}{x}\right)}^{6}·\left(\frac{1}{7}-\frac{7}{{x}^{2}}\right)$

$y=\text{tan}\phantom{\rule{0.1em}{0ex}}\left(\text{sec}\phantom{\rule{0.1em}{0ex}}x\right)$

$y=\text{csc}\phantom{\rule{0.1em}{0ex}}\left(\pi x+1\right)$

a. $f\left(u\right)=\text{csc}\phantom{\rule{0.1em}{0ex}}u,u=\pi x+1;$ b. $\text{−}\pi \phantom{\rule{0.1em}{0ex}}\text{csc}\phantom{\rule{0.1em}{0ex}}\left(\pi x+1\right)·\text{cot}\phantom{\rule{0.1em}{0ex}}\left(\pi x+1\right)$

$y={\text{cot}}^{2}x$

$y=-6\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{-3}x$

a. $f\left(u\right)=-6{u}^{-3},u=\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ b. $18\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{-4}x·\text{cos}\phantom{\rule{0.1em}{0ex}}x$

For the following exercises, find $\frac{dy}{dx}$ for each function.

$y={\left(3{x}^{2}+3x-1\right)}^{4}$

$y={\left(5-2x\right)}^{-2}$

$\frac{4}{{\left(5-2x\right)}^{3}}$

$y={\text{cos}}^{3}\left(\pi x\right)$

$y={\left(2{x}^{3}-{x}^{2}+6x+1\right)}^{3}$

$6{\left(2{x}^{3}-{x}^{2}+6x+1\right)}^{2}\left(3{x}^{2}-x+3\right)$

$y=\frac{1}{{\text{sin}}^{2}\left(x\right)}$

$y={\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x+\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)}^{-3}$

$-3{\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x+\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)}^{-4}·\left({\text{sec}}^{2}x+\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)$

$y={x}^{2}{\text{cos}}^{4}x$

$y=\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}7x\right)$

$-7\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}7x\right)·\text{sin}\phantom{\rule{0.1em}{0ex}}7x$

$y=\sqrt{6+\text{sec}\phantom{\rule{0.1em}{0ex}}\pi {x}^{2}}$

$y={\text{cot}}^{3}\left(4x+1\right)$

$-12\phantom{\rule{0.1em}{0ex}}{\text{cot}}^{2}\left(4x+1\right)·{\text{csc}}^{2}\left(4x+1\right)$

Let $y={\left[f\left(x\right)\right]}^{3}$ and suppose that ${f}^{\prime }\left(1\right)=4$ and $\frac{dy}{dx}=10$ for $x=1.$ Find $f\left(1\right).$

Let $y={\left(f\left(x\right)+5{x}^{2}\right)}^{4}$ and suppose that $f\left(-1\right)=-4$ and $\frac{dy}{dx}=3$ when $x=-1.$ Find ${f}^{\prime }\left(-1\right)$

$10\frac{3}{4}$

Let $y={\left(f\left(u\right)+3x\right)}^{2}$ and $u={x}^{3}-2x.$ If $f\left(4\right)=6$ and $\frac{dy}{dx}=18$ when $x=2,$ find ${f}^{\prime }\left(4\right).$

[T] Find the equation of the tangent line to $y=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\frac{x}{2}\right)$ at the origin. Use a calculator to graph the function and the tangent line together.

$y=\frac{-1}{2}x$

[T] Find the equation of the tangent line to $y={\left(3x+\frac{1}{x}\right)}^{2}$ at the point $\left(1,16\right).$ Use a calculator to graph the function and the tangent line together.

Find the $x$ -coordinates at which the tangent line to $y={\left(x-\frac{6}{x}\right)}^{8}$ is horizontal.

$x=±\sqrt{6}$

[T] Find an equation of the line that is normal to $g\left(\theta \right)={\text{sin}}^{2}\left(\pi \theta \right)$ at the point $\left(\frac{1}{4},\frac{1}{2}\right).$ Use a calculator to graph the function and the normal line together.

For the following exercises, use the information in the following table to find ${h}^{\prime }\left(a\right)$ at the given value for $a.$

$x$ $f\left(x\right)$ $f\prime \left(x\right)$ $g\left(x\right)$ $g\prime \left(x\right)$
0 2 5 0 2
1 1 −2 3 0
2 4 4 1 −1
3 3 −3 2 3

$h\left(x\right)=f\left(g\left(x\right)\right);a=0$

10

$h\left(x\right)=g\left(f\left(x\right)\right);a=0$

$h\left(x\right)={\left({x}^{4}+g\left(x\right)\right)}^{-2};a=1$

$-\frac{1}{8}$

$h\left(x\right)={\left(\frac{f\left(x\right)}{g\left(x\right)}\right)}^{2};a=3$

$h\left(x\right)=f\left(x+f\left(x\right)\right);a=1$

$-4$

$h\left(x\right)={\left(1+g\left(x\right)\right)}^{3};a=2$

$h\left(x\right)=g\left(2+f\left({x}^{2}\right)\right);a=1$

$-12$

$h\left(x\right)=f\left(g\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)\right);a=0$

[T] The position function of a freight train is given by $s\left(t\right)=100{\left(t+1\right)}^{-2},$ with $s$ in meters and $t$ in seconds. At time $t=6$ s, find the train’s

1. velocity and
2. acceleration.
3. Using a. and b. is the train speeding up or slowing down?

a. $-\frac{200}{343}$ m/s, b. $\frac{600}{2401}$ m/s 2 , c. The train is slowing down since velocity and acceleration have opposite signs.

[T] A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where $t$ is measured in seconds and $s$ is in inches:

$s\left(t\right)=-3\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\pi t+\frac{\pi }{4}\right).$

1. Determine the position of the spring at $t=1.5$ s.
2. Find the velocity of the spring at $t=1.5$ s.

[T] The total cost to produce $x$ boxes of Thin Mint Girl Scout cookies is $C$ dollars, where $C=0.0001{x}^{3}-0.02{x}^{2}+3x+300.$ In $t$ weeks production is estimated to be $x=1600+100t$ boxes.

1. Find the marginal cost ${C}^{\prime }\left(x\right).$
2. Use Leibniz’s notation for the chain rule, $\frac{dC}{dt}=\frac{dC}{dx}·\frac{dx}{dt},$ to find the rate with respect to time $t$ that the cost is changing.
3. Use b. to determine how fast costs are increasing when $t=2$ weeks. Include units with the answer.

a. ${C}^{\prime }\left(x\right)=0.0003{x}^{2}-0.04x+3$ b. $\frac{dC}{dt}=100·\left(0.0003{x}^{2}-0.04x+3\right)$ c. Approximately \$90,300 per week

[T] The formula for the area of a circle is $A=\pi {r}^{2},$ where $r$ is the radius of the circle. Suppose a circle is expanding, meaning that both the area $A$ and the radius $r$ (in inches) are expanding.

1. Suppose $r=2-\frac{100}{{\left(t+7\right)}^{2}}$ where $t$ is time in seconds. Use the chain rule $\frac{dA}{dt}=\frac{dA}{dr}·\frac{dr}{dt}$ to find the rate at which the area is expanding.
2. Use a. to find the rate at which the area is expanding at $t=4$ s.

[T] The formula for the volume of a sphere is $S=\frac{4}{3}\pi {r}^{3},$ where $r$ (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun.

1. Suppose $r=\frac{1}{{\left(t+1\right)}^{2}}-\frac{1}{12}$ where $t$ is time in minutes. Use the chain rule $\frac{dS}{dt}=\frac{dS}{dr}·\frac{dr}{dt}$ to find the rate at which the snowball is melting.
2. Use a. to find the rate at which the volume is changing at $t=1$ min.

a. $\frac{dS}{dt}=-\frac{8\pi {r}^{2}}{{\left(t+1\right)}^{3}}$ b. The volume is decreasing at a rate of $-\frac{\pi }{36}$ ft 3 /min.

[T] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function $T\left(x\right)=94-10\phantom{\rule{0.1em}{0ex}}\text{cos}\left[\frac{\pi }{12}\left(x-2\right)\right],$ where $x$ is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.

[T] The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function $D\left(t\right)=5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\frac{\pi }{6}t-\frac{7\pi }{6}\right)+8,$ where $t$ is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.

$~2.3$ ft/hr

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