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.)

Differentiating a Composite Function

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.

Differentiating a Composite Function

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.

Finding an Equation of a Tangent Line

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.

Application to Carbon-14 Dating

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.

1. What is the rate of change of the percentage of carbon-14 present in a \(550\)-year-old fossil?
2. What is the rate of change of the percentage of carbon-14 present in a \(2000\)-year-old fossil?

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.

Differentiating a Composite Function

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 \[\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime = \dfrac{d}{dx}a^{x}=a^{x}\ln a} \]

Differentiating Exponential Functions

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

Using the Power Rule for Functions to Find a Derivative

1. If \(f( x) =(3-x^{3}) ^{-5}\), then \[ \begin{eqnarray*} f^\prime ( x) &=&\dfrac{d}{dx}(3-x^{3}) ^{-5} \underset{\underset{\color{#0066A7}{\hbox{for Functions}}}{\underset{\color{#0066A7}{\hbox{Power Rule}}}{{\color{#0066A7}{\uparrow}}}}}{=} -5(3-x^{3}) ^{-5-1}\cdot \dfrac{d}{dx}(3-x^{3}) \\ &=& -5(3-x^{3}) ^{-6}\cdot ( -3x^{2}) = 15x^{2}(3-x^{3}) ^{-6}=\dfrac{15x^{2}}{( 3-x^{3}) ^{6}} \end{eqnarray*} \]
2. If \(f( \theta ) =\cos ^{3}\theta \), then \(f( \theta ) =( \cos \theta ) ^{3},\) and \[ \begin{eqnarray*} f^\prime ( \theta ) &=&\dfrac{d}{d\theta }( \cos \theta ) ^{3} \underset{\underset{\color{#0066A7}{\hbox{for Functions}}}{\underset{\color{#0066A7}{\hbox{Power Rule}}}{{\color{#0066A7}{\uparrow}}}}}{=} 3( \cos \theta ) ^{3-1}\cdot \dfrac{d}{ d\theta }\cos \theta =3\cos ^{2}\theta \cdot ( -\sin \theta)\\ &=&-3\cos ^{2}\theta \sin \theta \end{eqnarray*} \]

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.

Using the Power Rule for Functions with Other Derivative Rules

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.

Differentiating a Composite Function

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.