OBJECTIVES

When you finish this section, you should be able to:

1. Differentiate a composite function (p. 198)
2. Differentiate ${y=a^{x}}$, ${a>0}$, ${a≠ 1}$ (p. 202)
3. Use the Power Rule for functions to find a derivative (p. 202)
4. Use the Chain Rule for multiple composite functions (p. 204)

NEED TO REVIEW?

Composite functions and their properties are discussed in Section P.3, pp. 25-27.

IN WORDS

If you think of the function $y= f(u)$ as the outside function and the function $u= g(x)$ as the inside function, then the derivative of ${f\circ g}$ is the derivative of the outside function, evaluated at the inside function, times the derivative of the inside function.
That is,

THEOREM Chain Rule

If a function $g$ is differentiable at $x_{0}$ and a function $f$ is differentiable at $g( x_{0}) ,$ then the composite function $f\circ g$ is differentiable at $x_{0}$ and $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ (f\circ g)^\prime ( x_{0}) =f^\prime (g( x_{0})) \cdot g^\prime ( x_{0})}}$$

For differentiable functions $y=f( u)$ and $u=g( x)$, the Chain Rule, in Leibniz notation, takes the form $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}}$$ where in $\dfrac{dy}{du}$ we substitute $u=g( x)$.

Partial Proof

The Chain Rule is proved using the definition of a derivative. First we observe that if $x$ changes by a small amount $\Delta x$, the corresponding change in $u=g( x)$ is $\Delta u$. That is, $\Delta u$ depends on $\Delta x$. Also, $$g^\prime ( x) =\dfrac{du}{dx}=\lim\limits_{\Delta x\rightarrow 0} \dfrac{\Delta u}{\Delta x}$$

Since $y=f( u)$, the change $\Delta u$, which could equal 0, causes a change $\Delta y$ . If $\Delta u$ is never $0,$ then $$f^\prime ( u) =\dfrac{dy}{du}=\lim\limits_{\Delta u\rightarrow 0} \dfrac{\Delta y}{\Delta u}$$

To find $\dfrac{dy}{dx}$, we write $$\begin{eqnarray*} \dfrac{dy}{dx} &= &\lim\limits_{\Delta x\rightarrow 0}\dfrac{\Delta y}{\Delta x} \underset{\underset{\color{#0066A7}{{\hbox{\(\Delta u≠ 0\)}}}}{\color{#0066A7}{\uparrow }}}{=} \lim\limits_{\Delta x\rightarrow 0}\left( \dfrac{\Delta y}{\Delta x}\cdot \dfrac{\Delta u}{ \Delta u}\right)= \lim\limits_{\Delta x\rightarrow 0}\left( \dfrac{\Delta y}{ \Delta u}\cdot \dfrac{\Delta u}{\Delta x}\right) \\ &=& \left( \lim\limits_{\Delta x\rightarrow 0}\dfrac{\Delta y}{\Delta u}\right) \left( \lim\limits_{\Delta x\rightarrow 0}\dfrac{\Delta u}{\Delta x}\right) \end{eqnarray*}$$

Since the differentiable function $u=g( x)$ is continuous, $\Delta u\rightarrow 0$ as $\Delta x\rightarrow 0,$ so in the first factor we can replace $\Delta x\rightarrow 0$ by $\Delta u\rightarrow 0$. Then $$\dfrac{dy}{dx}=\left( \lim\limits_{\Delta u\rightarrow 0}\dfrac{\Delta y}{ \Delta u}\right) \left( \lim\limits_{\Delta x\rightarrow 0}\dfrac{\Delta u}{ \Delta x}\right) =\dfrac{dy}{du}\cdot \dfrac{du}{dx}$$

This proves the Chain Rule if $\Delta u$ is never $0$. To complete the proof, we need to consider the case when $\Delta u$ may be $0$. (This part of the proof is given in Appendix B.)

Find the derivative of:
(a) $y=( x^{3}-4x+1) ^{100}$ $\quad$ (b) $y=\cos \left( 3x-\dfrac{\pi}{4}\right)$

Solution (a) In the composite function $y=( x^{3}-4x+1)^{100},$ let $u=x^{3}-4x+1.$ Then $y=u^{100}.$ Now $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$ are $$\begin{eqnarray*} \dfrac{dy}{du}=\dfrac{d}{du}u^{100}=100u^{99}\underset{\underset{\color{#0066A7}{\hbox {\(u=x^{3}-4x+1\)}}}{\color{#0066A7}{\uparrow}}}{=}100(x^{3}-4x+1)^{99} \end{eqnarray*}$$

and $$\begin{eqnarray*} \dfrac{du}{dx}=\dfrac{d}{dx}( x^{3}-4x+1) =3x^{2}-4 \end{eqnarray*}$$

We use the Chain Rule to find $\dfrac{dy}{dx}.$ $$\begin{eqnarray*} && \dfrac{dy}{dx}\underset{\underset {\color{#0066A7} {\hbox{ Chain Rule}}} {\color{#0066A7}{{\uparrow}}}}{=}\dfrac{dy}{du}\cdot \dfrac{du}{dx}=100( x^{3}-4x+1) ^{99}(3x^{2}-4)\\ \end{eqnarray*}$$

(b) In the composite function $y=\cos \left( 3x-\dfrac{\pi }{4} \right)$, let $u=3x-\dfrac{\pi }{4}.$ Then $y=\cos u$ and $$\begin{eqnarray*} && \dfrac{dy}{du}=\dfrac{d}{du}\cos u=-\sin u \underset{\underset{\color{#0066A7} {\hbox{\(u=3x-{\dfrac{\pi}{4}}\)}}}{\color{#0066A7}{{\uparrow}}}}{=} -\sin \left( 3x-\dfrac{\pi}{4}\right) \quad\hbox{and}\quad \dfrac{du}{dx}=\dfrac{d}{dx}\left( 3x-\dfrac{\pi }{4}\right) =3\\ \end{eqnarray*}$$

Now we use the Chain Rule. $$\begin{eqnarray*} && \dfrac{dy}{dx}\underset{\underset{\color{#0066A7}{\hbox{Chain Rule}}}{\color{#0066A7}{{\uparrow}}}}{=}\dfrac{dy}{du}\cdot \dfrac{du}{dx}=-\!\sin\! \left(3x-\dfrac{\pi }{4}\right) \cdot 3=-3\sin \left(3x-\dfrac{\pi }{4}\right)\\ \end{eqnarray*}$$

NOW WORK

Problems 9 and 37.

Find $y^\prime$ if:

(a) $y=e^{x^{2}-4}$ $\quad$ (b) $y=\sin (4e^{x})$

Solution (a) For $y=e^{x^{2}-4},$ we let $u=x^{2}-4.$ Then $y=e^{u}$ and $$\begin{eqnarray*} && \dfrac{dy}{du}=\dfrac{d}{du}e^{u}=e^{u}\underset{\underset{\color{#0066A7}{{{\hbox{\(u=x^{2}-4\)}}}}} {\color{#0066A7}{\uparrow}}}{=}e^{x^{2}-4} \quad\hbox{and}\quad \dfrac{du}{dx}=\dfrac{d}{dx}( x^{2}-4) =2x\\ \end{eqnarray*}$$

Using the Chain Rule, we get $$y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=e^{x^{2}-4}\cdot 2x=2xe^{x^{2}-4}$$

(b) For $y=\sin (4e^{x})$, we let $u=4e^{x}$. Then $y=\sin u$ and $$\begin{eqnarray*} && \dfrac{dy}{du}=\dfrac{d}{du}\sin u=\cos u \underset{\underset{\color{#0066A7}{{{\hbox{\(u=4e^{x}\)}}}}}{\color{#0066A7}{\uparrow}}}{=}\cos (4e^{x})\quad\hbox{and}\quad\dfrac{du}{dx}=\dfrac{d}{dx}(4e^{x})=4e^{x}\\ \end{eqnarray*}$$

Using the Chain Rule, we get $$y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\cos (4e^{x}) \cdot 4e^{x}=4e^{x}\cos (4e^{x})$$

NOW WORK

Problem 41.

Find an equation of the tangent line to the graph of $y=5e^{4x}$ at the point $( 0,5) .$

Figure 1

Solution The slope of the tangent line to the graph of $y=f( x)$ at the point $( 0,5)$ is $f^\prime ( 0)$. $$\begin{eqnarray*} f^\prime (x) &=& \dfrac{d}{dx}( 5e^{4x}) \underset{\underset{\color{#0066A7}{\hbox{Mulitple Rule}}}{\underset{\color{#0066A7}{\hbox{Constant}}}{\color{#0066A7}{{\uparrow}}}}}{=} 5\dfrac{d}{dx}e^{4x} \underset{\underset{\underset{\color{#0066A7}{{{\hbox{\(\tfrac{d}{dx}{e}^{u}{= e}^{u} \tfrac{du}{dx}\)}}}}}{\color{#0066A7}{{{\hbox{\(u=4x;\)}}}}}}{\color{#0066A7} {{\uparrow}}}}{=} 5e^{4x}\,{\cdot}\,\dfrac{d}{dx}(4x)=5e^{4x}\,{\cdot}\,4=20e^{4x}\\ \end{eqnarray*}$$ $m_{\tan }=$ $f^\prime ( 0) =20e^{0}=20$. Using the point slope form of a line, we have $$\begin{eqnarray*} y-5 &=&20( x-0)\qquad {\color{#0066A7}{y-y_{0} = m_{\tan }( { x-x}_{{ 0}})}}\\ y &=&20x+5 \end{eqnarray*}$$

NOW WORK

Problem 77.

All carbon on Earth contains some carbon-14, which is radioactive. When a living organism dies, the carbon-14 begins to decay at a fixed rate. The formula $P( t) =100e^{-0.000121t}$ gives the percentage of carbon-14 present at time $t$ years. Notice that when $t=0$, the percentage of carbon-14 present is 100%. When the preserved bodies of $15$-year-old La Doncella and her two children were found in Peru in 2005, 93.5% of the carbon-14 remained in their bodies, indicating that the three had died about $550$ years earlier.

Solution (a) The rate of change of $P$ is given by its derivative $$\begin{eqnarray*} && P^\prime ( t) =\dfrac{d}{dt}(100e^{-0.000121t}) \underset{\underset{\color{#0066A7} { {\hbox{\( \tfrac{d}{dt}e^{u}=e^{u}\tfrac{du}{dt} \)}} }}{\color{#0066A7}{{\uparrow}}}}{=} 100(-0.000121e^{-0.000121t}) = -0.0121e^{-0.000121t}\\ \end{eqnarray*}$$

At $t=550$ years, $$P^\prime (550) =-0.0121e^{-0.000121(550)} \approx -0.0113$$

The percentage of carbon-14 present in a $550$-year-old fossil is decreasing at the rate of 1.13% per year.

(b) When $t=2000$ years, the rate of change is $$P^\prime ( 2000) =-0.0121e^{-0.000121( 2000) }\approx -0.0095$$

The percentage of carbon-14 present in a $2000$-year-old fossil is decreasing at the rate of 0.95% per year.

NOW WORK

Problem 99.

Suppose $h=f\circ g$. Find $h^\prime (1)$ given that: $$f(1) =2 \quad f^\prime (1) =1 \quad f^\prime (2) =-4 \quad g(1) =2 \quad g^\prime (1) =-3\quad g^\prime (2) =5$$

Solution Based on the Chain Rule using prime notation, we have $$h^\prime ( x_{0}) =( f\circ g)^\prime ( x_{0}) =f^\prime (g( x_{0})) g^\prime ( x_{0})$$

When $x_{0}=1,$ $$\begin{eqnarray*} && h^\prime (1) =f^\prime (g(1)) g^\prime(1) \underset{\underset{\color{#0066A7}{{g(1)=2; g^\prime (1)=-3}}}{\color{#0066A7}{{\uparrow}}}}{=} f^\prime (2) \cdot ( -3) \underset{\underset{\color{#0066A7}{{f^\prime (2) =-4}}} {\color{#0066A7}{{\uparrow}}}}{=}(-4) ( -3) =12 \\ \end{eqnarray*}$$

NOW WORK

Problem 81.

NEED TO REVIEW?

Properties of logarithms are discussed in Appendix A.1, pp. A-10 to A-11.

THEOREM Derivative of ${y=a^x}$

The derivative of an exponential function $y=a^{x}$, where $a>0$ and $a≠ 1,$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime = \dfrac{d}{dx}a^{x}=a^{x}\ln a}$$

Find the derivative of each function:
(a) $f( x) =2^{x}$ $\quad$ (b) $F( x) =3^{-x}$ $\quad$ (c) $g( x) =\left(\dfrac{1}{2}\right) ^{x^{2}+1}$

Solution (a) $f$ is an exponential function with base $a=2$. $$\begin{eqnarray*} f^\prime (x) =\dfrac{d}{dx}2^{x}= 2^{x}\ln 2\qquad {\color{#0066A7}{\tfrac{d}{dx}a^{x}=a^{x}\ln a}} \end{eqnarray*}$$

(b) Since $F(x)=3^{-x}=\dfrac{1}{3^x}=\left( \dfrac{1}{3}\right) ^{x}$, $F$ is an exponential function with base $\dfrac{1}{3}.$ So, $$\begin{eqnarray*} && F^\prime (x) =\dfrac{d}{dx}\left( \dfrac{1}{3}\right) ^{x} \underset{\underset{{\color{#0066A7}{\tfrac{d}{dx}a^{x}=a^{x}\ln a}}}{\color{#0066A7}{{\uparrow}}}}{=} \left( \dfrac{1}{3}\right) ^{x}\ln \dfrac{1}{3}=\left( \dfrac{1}{3}\right) ^{x}\ln 3^{-1}=-\left( \dfrac{1}{3}\right) ^{x}\ln 3 =-\dfrac{1}{3^{x}}\ln 3 \\ \end{eqnarray*}$$

(c) $y=g( x) =\left( \dfrac{1}{2}\right) ^{x^{2}+1}$ is a composite function. If $u=x^{2}+1,$ then $y=\left( \dfrac{1}{2}\right) ^{u}$ and $$\begin{eqnarray*} && \dfrac{dy}{du} \underset{\underset{{\color{#0066A7}{\tfrac{d}{du}a^{u}=a^{u}\ln a}}}{{\color{#0066A7}{\uparrow}}}}{=} \left(\dfrac{1}{2}\right)^{u}\ln \left( \dfrac{1}{2}\right) \underset{\underset{{\color{#0066A7}{\ln \left( \tfrac{1}{2}\right) =-\ln 2}}}{{\color{#0066A7}{\uparrow}}}}{=} -\left( \dfrac{1}{2}\right) ^{u}\ln 2 \underset{\underset{{\color{#0066A7}{u=x^{2}+1}}}{{\color{#0066A7}{\uparrow}}}}{=} -\left( \dfrac{1}{2}\right) ^{x^{2}+1}\ln 2 \quad \hbox{and}\quad \dfrac{du}{dx}=2x\\ \end{eqnarray*}$$ So, by the Chain Rule, $$\begin{eqnarray*} g^\prime (x) &=&\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\left[ -\!\left( \dfrac{1}{2}\right) ^{x^{2}+1}\ln 2\right] (2x) =(-\ln 2)\,{x}\,\!\left( \dfrac{1}{2}\right) ^{x^{2}} \end{eqnarray*}$$

NOW WORK

Problem 47.

THEOREM Power Rule for Functions

If $g$ is a differentiable function and $n$ is an integer, then $$\bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{d}{dx}[g( x) ] ^{n}=n[g(x)] ^{n-1}g^\prime (x)}$$

Proof

If $y=[g( x) ] ^{n},$ let $y=u^{n}$ and $u=g( x) .$ Then $$\dfrac{dy}{du}=nu^{n-1}=n[g( x) ] ^{n-1}\qquad \hbox{and} \qquad \dfrac{du}{dx}=g^\prime (x)$$

By the Chain Rule, $$y^\prime =\dfrac{d}{dx}[g( x) ] ^{n}=\dfrac{dy}{du} \cdot \dfrac{du}{dx}=n[g( x) ] ^{n-1}g^\prime ( x)$$

NOW REWORK

Example 7 using the Chain Rule and rework Example 1(a) using the Power Rule for Functions. Decide for yourself which method is easier.

NOW WORK

Problem 21.

Find the derivative of:
(a) $f( x) =e^{x}(x^{2}+1) ^{3}$ $\quad$ (b) $g( x) =\left( \dfrac{3x+2}{4x^{2}-5}\right)^{5}$

Solution (a) The function $f$ is the product of $e^{x}$ and $(x^{2}+1) ^{3}$, so we first use the Product Rule. $$\begin{eqnarray*} && f^\prime (x) \underset{\underset{\color{#0066A7}{\hbox{Product Rule}}}{{\color{#0066A7}{\uparrow}}}}{=} e^{x}\left[ \dfrac{d}{dx}(x^{2}+1) ^{3}\right] +\left[ \dfrac{d}{dx}e^{x}\right] (x^{2}+1) ^{3}\\ \end{eqnarray*}$$

To complete the solution, we use the Power Rule for Functions to find $\dfrac{d}{dx}(x^{2}+1) ^{3}$. $$\begin{eqnarray*} f^\prime ( x) &=& e^{x}\left[ 3(x^{2}+1) ^{2}\cdot \dfrac{d}{dx}(x^{2}+1) \right] +e^{x}(x^{2}+1) ^{3}\qquad {{\color{#0066A7}{\hbox{Power Rule for Functions}}}} \\ &=& e^{x}[3(x^{2}+1) ^{2}\cdot 2x] + e^{x}(x^{2}+1) ^{3} = e^{x}[6x(x^{2}+1) ^{2}+(x^{2}+1) ^{3}]\\ &=& e^{x}(x^{2}+1) ^{2}[6x+x^{2}+1] =e^{x}(x^{2}+1) ^{2}( x^{2}+6x+1) \end{eqnarray*}$$

(b) $g$ is a function raised to a power, so we begin with the Power Rule for Functions. $$\begin{eqnarray*} g^\prime ( x) &=&\dfrac{d}{dx}\left( \dfrac{3x+2}{4x^{2}-5}\right) ^{5} = 5\left( \dfrac{3x+2}{4x^{2}-5}\right) ^{4}\left[ \dfrac{d}{dx}\left( \dfrac{3x+2}{4x^{2}-5}\right) \right] \,\, {{\color{#0066A7}{\hbox{Power Rule for Functions}}}}\\ &=&5\left( \dfrac{3x+2}{4x^{2}-5}\right) ^{4}\left[ \dfrac{( 3) ( 4x^{2}-5) -(3x+2) (8x) }{(4x^{2}-5) ^{2}}\right]\quad \qquad {{\color{#0066A7}{\hbox{Quotient Rule}}}}\\ &=&\dfrac{5(3x+2) ^{4}[ (12x^{2}-15) -( 24x^{2}+16x) ] }{(4x^{2}-5) ^{6}}\\ &=&\dfrac{5(3x+2) ^{4}[ -12x^{2}-16x-15] }{( 4x^{2}-5) ^{6}} \end{eqnarray*}$$

NOW WORK

Problem 29.

Find $y^\prime$ if:
(a) $y=5\cos ^{2}(3x+2)$ $\quad$ (b) $y=\sin ^{3}\left( \dfrac{\pi }{2}x\right)$

Solution (a) For $y=5\cos ^{2}(3x+2) ,$ we use the Chain Rule with $y=5u^{2}$, $u=\cos v$, and $v=3x+2.$ Then $y=5u^{2}=5\cos ^{2}v=5\cos ^{2}(3x+2)$ and $$\begin{eqnarray*} \dfrac{dy}{du} &=&\dfrac{d}{du}(5u^{2}) =10u \underset{\underset{\color{#0066A7}{{\hbox{\(v=3x+2\)}}}}{\underset{\color{#0066A7}{{\hbox{\(u=cos v\)}}}}{{\color{#0066A7}{\uparrow}}}}} {=} 10\cos (3x+2) \\ \dfrac{du}{dv} &=&\dfrac{d}{dv}\cos v=-\sin v \underset{\underset{{\color{#0066A7}{{\hbox{\(v=3x+2\)}}}}}{{\color{#0066A7}{\uparrow}}}}{=} -\sin (3x+2) \\ \dfrac{dv}{dx} &=&\dfrac{d}{dx}(3x+2) =3 \end{eqnarray*}$$

Then $$\begin{eqnarray*} y^\prime &=& \dfrac{dy}{dx} \underset{\underset{\color{#0066A7}{\scriptsize\hbox{Chain Rule}}}{{\color{#0066A7}{\uparrow}}}}{=} \dfrac{dy}{du}\cdot \dfrac{du}{dv}\cdot \dfrac{dv}{dx}=[10\cos (3x+2)] [ -\!\sin (3x+2)] [ 3]\\ &=& -30\cos (3x+2) \sin (3x+2) \end{eqnarray*}$$

(b) For $y=\sin ^{3}\left( \dfrac{\pi }{2}x\right) ,$ we use the Chain Rule with $y=u^{3}$, $u=\sin v\,$, and $v=\dfrac{\pi }{2}x$. Then $y=u^{3}=\left( \sin v\right) ^{3}=\left[ \sin \left( \dfrac{\pi }{2}x\right) \right] ^{3}=\sin ^{3}\left( \dfrac{\pi }{2}x\right)$, and $$\begin{eqnarray*} \dfrac{dy}{du} &=&\dfrac{d}{du}u^{3}=3u^{2} \underset{\underset{\underset{\color{#0066A7}{{\hbox{\(v=\dfrac{\pi}{2}x\)}}}}{\color{#0066A7}{{\hbox{\(u=\sin v\)}}}}}{{\color{#0066A7}{\uparrow}}}}{=} 3\left[ \sin \left( \dfrac{\pi }{2}x\right) \right] ^{2} = 3 \sin^{2}\left(\frac{\pi}{2}x\right)\\ \dfrac{du}{dv} &=&\dfrac{d}{dv}\sin v=\cos v \underset{\underset{\color{#0066A7}{{\hbox{\(v=\dfrac{\pi}{2}x\)}}}}{{\color{#0066A7}{\uparrow}}}}{=}\cos \left( \dfrac{\pi}{2}x\right) \\ \dfrac{dv}{dx} &=&\dfrac{d}{dx}\left( \dfrac{\pi }{2}x\right) = \dfrac{\pi }{2} \end{eqnarray*}$$

Then $$\begin{eqnarray*} y^\prime &=& \dfrac{dy}{dx} \underset{\underset{\color{#0066A7}{\scriptsize\hbox{Chain Rule}}}{{\color{#0066A7}{\uparrow}}}}{=} \dfrac{dy}{du}\cdot \dfrac{du}{dv}\cdot \dfrac{dv}{dx}=3\sin ^{2}\left( \dfrac{\pi }{2}x\right) \cdot \cos \left( \dfrac{\pi }{2}x\right) \cdot \left( \dfrac{\pi }{2}\right) \\ &=&\dfrac{3\pi }{2}\sin ^{2}\!\left( \dfrac{\pi }{2}x\right) \cos \left( \dfrac{\pi }{2}x\right) \\ \end{eqnarray*}$$

NOW WORK

Problem 59.