OBJECTIVES

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

1. Differentiate the product of two functions (p. 173)
2. Differentiate the quotient of two functions (p. 175)
3. Find higher-order derivatives (p. 177)
4. Work with acceleration (p. 179)

IN WORDS

The derivative of the product of two differentiable functions equals the first function times the derivative of the second function plus the derivative of the first function times the second function. That is, $(fg)' =f (g^\prime ) + ( f^\prime ) g$.

THEOREM Product Rule

If $f$ and $g$ are differentiable functions and if $F(x) =f(x) g(x)$, then $F$ is differentiable, and the derivative of the product $F$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt] {F' (x) =[f(x) g(x)]' =f(x) g' (x) +f' (x) g(x)}}$$

In Leibniz notation, the Product Rule takes the form $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt] {\dfrac{d}{dx}F( x) =\dfrac{d}{dx}[ f( x) g( x) ] =f( x) \left[ \dfrac{d}{ dx}g( x) \right] +\left[ \dfrac{d}{dx}f( x) \right] g( x)}}$$

Find $y^\prime$ if $y=( 1+x^{2}) e^{x}\!$.

Solution The function $y$ is the product of two functions: a polynomial, $f( x) =1+x^{2}$, and the exponential function $g( x) =e^{x}$. By the Product Rule, $$\begin{eqnarray*} &&y^\prime =\frac{d}{dx}[(1+x^{2}) e^{x}] {=} (1+x^{2}) \left[ \frac{d}{dx}e^{x}\right] +\left[\frac{d}{dx}(1+x^{2})\right] e^{x}=( 1+x^{2}) e^{x}+2xe^{x}\\[-6.4pt] &&\hspace{4.9pc}\underset{\color{#0066A7}{\scriptsize \hbox{Product Rule}}}{\color{#0066A7}{\uparrow}} \end{eqnarray*}$$

At this point, we have found the derivative, but it is customary to simplify the answer. Then $$\begin{eqnarray*} &&y\prime {=} ( 1+x^{2}+2x) e^{x} {=} ( x+1) ^{2}e^{x}\\[-6.4pt] &&\hspace{-0.7pc}\underset{\color{#0066A7}{\scriptsize \hbox{Factor out \(e^{x}\).}}}{\color{#0066A7}{\uparrow }}\hspace{2.5pc}\underset{\color{#0066A7}{\scriptsize \hbox{Factor}.}}{\color{#0066A7}{\uparrow }} \end{eqnarray*}$$

NOW WORK

Problem 9.

Find the derivative of $F(v)=( 5v^{2}-v+1) ( v^{3}-1)$ in two ways:

Solution

(a) $F$ is the product of the two functions $f(v) =5v^{2}-v+1$ and $g( v) =v^{3}-1$. Using the Product Rule, we get $$\begin{eqnarray*} F^\prime ( v) &=& ( 5v^{2}-v+1) \left[\dfrac{d}{dv}(v^{3}-1)\right]+\left[\dfrac{d}{dv}(5v^{2}-v+1)\right]( v^{3}-1)\\[5pt] &=&(5v^{2}-v+1) (3v^{2}) +(10v-1) (v^{3}-1)\\[5pt] &=&15v^{4}-3v^{3}+3v^{2}+10v^{4}-10v-v^{3}+1\\[5pt] &=&25v^{4}-4v^{3}+3v^{2}-10v+1 \end{eqnarray*}$$

(b) Here, we multiply the factors of $F$ before differentiating. $$F\left( v\right) =( 5v^{2}-v+1) ( v^{3}-1) =5v^{5}-v^{4}+v^{3}-5v^{2}+v-1$$

Then $$F' ( v) =25v^{4}-4v^{3}+3v^{2}-10v+1$$

NOW WORK

Problem 13.

THEOREM Quotient Rule

If two functions $f$ and $g$ are differentiable and if $F(x) =\dfrac{f(x) }{g(x) }$, $g(x)\neq 0$, then $F$ is differentiable, and the derivative of the quotient $F$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt] F' (x) =\left[ \dfrac{f(x) }{g(x) }\right] ^{\prime }=\dfrac{f^\prime (x) g(x) -f(x) g^\prime ( x) }{\left[ g( x) \right] ^{2}} }$$

In Leibniz notation, the Quotient Rule takes the form $$\bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{d}{dx}F( x) = \dfrac{d}{dx}\left[ \dfrac{f( x) }{g( x) }\right] =\dfrac{\left[ \dfrac{d}{ dx}f( x) \right] g( x) -f( x) \left[ \dfrac{d}{dx}g( x) \right] }{[ g( x) ] ^{2}}}$$

PROOF

We use the definition of a derivative to find $F' (x)$. $$\begin{eqnarray*} F' ( x) &=&\lim\limits_{h\rightarrow 0}\dfrac{F( x+h) -F( x) }{h} \underset{\underset{\color{#0066A7}{F( x) =\dfrac{f(x) }{g(x) }}}{\color{#0066A7}{\uparrow }}}{=} \lim\limits_{h \rightarrow 0}\dfrac{\dfrac{f( x+h) }{g( x+h) }-\dfrac{ f(x) }{g( x) }}{h}\\ &=&\lim\limits_{h\rightarrow 0}\dfrac{ f(x+h) g(x) -f(x) g(x+h) }{ h[g( x+h) g( x) ] } \end{eqnarray*}$$

We write $F’$ in an equivalent form that contains the difference quotients for $f$ and $g$ by subtracting and adding $f(x) g(x)$ to the numerator. $$F' ( x) =\lim\limits_{h\rightarrow 0}\dfrac{f( x+h) g( x) -f( x) g( x) +f( x) g( x) -f( x) g( x+h) }{h[ g( x+h) g( x) ] }$$

Now we group and factor the numerator. $$\begin{eqnarray*} F' ( x) &=&\lim\limits_{h\rightarrow 0}\dfrac{[ f( x+h) -f( x) ] g( x) -f( x) [ g( x+h) -g( x) ] }{h[ g( x+h) g( x) ] }\\[5pt] \notag \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{\left[ \dfrac{f( x+h) -f( x) }{h}\right] g( x) -f( x) \left[ \dfrac{g( x+h) -g( x) }{h}\right] }{g( x+h) g( x) }\\[5pt] \notag \\ &=&\dfrac{\lim\limits_{h\rightarrow 0}\left[ \dfrac{f( x+h) -f( x) }{h}\right] \cdot \lim\limits_{h\rightarrow 0}g( x) -\lim\limits_{h\rightarrow 0}f( x) \cdot \lim\limits_{h\rightarrow 0}\left[ \dfrac{g( x+h) -g( x) }{h}\right] }{\lim\limits_{h\rightarrow 0}g( x+h) \cdot \lim\limits_{h\rightarrow 0}g( x) }\\[5pt] \notag \\ &=&\dfrac{f^\prime ( x) g( x) -f( x) g^\prime ( x) }{[ g( x) ] ^{2}} \end{eqnarray*}$$

RECALL

Since $g$ is differentiable, it is continuous; so, $\lim\limits_{h\rightarrow 0}{ g} ( x+h ) =g (x) .$

Find $y^\prime$ if $y=\dfrac{x^{2}+1}{2x-3}$.

Solution The function $y$ is the quotient of $f( x) =x^{2}+1$ and $g( x) =2x-3$. Using the Quotient Rule, we have $$\begin{eqnarray*} y^\prime &=&\dfrac{d}{dx}\!\left(\! \dfrac{x^{2}+1}{2x-3}\!\right) =\dfrac{\left[ \dfrac{d}{dx}( x^{2}+1) \right]\! ( 2x-3) -(x^{2}+1) \left[ \dfrac{d}{dx} \! ( 2x-3) \right] }{( 2x-3) ^{2}} \\[5pt] &=&\dfrac{( 2x) ( 2x-3) -( x^{2}+1) (2) }{( 2x-3) ^{2}}=\dfrac{4x^{2}-6x-2x^{2}-2}{( 2x-3) ^{2}}=\dfrac{2x^{2}-6x-2}{( 2x-3) ^{2}} \end{eqnarray*}$$

provided $x\neq \dfrac{3}{2}$.

NOW WORK

Problem 23.

COROLLARY Derivative of the Reciprocal of a Function

If a function $g$ is differentiable, then $$\bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{d}{dx}\!\left[ \dfrac{1}{g( x) }\right] =-\dfrac{ \dfrac{d}{dx}g( x) }{\left[ g( x) \right] ^{2}}=- \dfrac{g^\prime ( x) }{\left[ g( x) \right] ^{2}} }\tag{1}$$

provided $g( x) \neq 0$.

NOW WORK

Problem 25.

THEOREM Power Rule

The derivative of $y=x^{n},$ where $n$ any integer, is $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt] y^\prime =\dfrac{d}{dx}x^{n}=nx^{n-1} }$$

NOW WORK

Problem 31.

Ohm’s Law states that the current $I$ running through a wire is inversely proportional to the resistance $R$ in the wire and can be written as $I=\dfrac{ V}{R}$, where $V$ is the voltage. Find the rate of change of $I$ with respect to $R$ when $V$=12 volts.

Solution The rate of change of $I$ with respect to $R$ is the derivative $\dfrac{dI}{dR}$. We write Ohm’s Law with $V=12$ as $I=\dfrac{V}{R }=12R^{-1}$ and use the Power Rule. $$\dfrac{dI}{dR}=\dfrac{d}{dR}( 12R^{-1}) =12\cdot \dfrac{d}{dR} R^{-1}=12( -1R^{-2}) =-\dfrac{12}{R^{2}}$$

The minus sign in $\dfrac{dI}{dR}$ indicates that the current $I$ decreases as the resistance $R$ in the wire increases.

NOW WORK

Problem 85.

Find the second, third, and fourth derivatives of $y=2x^{3}$.

Solution Since $y$ is a power function, we use the Simple Power Rule and the Constant Multiple Rule to find each derivative. The first derivative is $$y^\prime =\dfrac{d}{dx}( 2x^{3}) = 2\cdot \dfrac{d}{dx}x^{3}=2\cdot 3x^{2}=6x^{2}$$

The next three derivatives are

NOTE

If $n >1$ is an integer, the product $n\cdot ( n-1 ) \cdot ( n-2 ) \cdot \cdots \cdot 3\cdot 2\cdot 1$ is often written $n!$ and is read, “n factorial.” The factorial symbol $n!$ means 0!=1, 1!=1, and $n!=1\cdot 2\cdot 3\cdot \cdots \cdot ( n-1 ) \cdot n$, where $n>1$.

NOW WORK

Problem 41.

Find the second and third derivatives of $y=( 1+x^{2})e^{x}$.

Solution In Example 1, we found that $y^\prime =(1+x^{2}) e^{x}+2xe^{x}=(x^{2}+2x+1) e^{x}$. To find $y''$, we use the Product Rule with $y^\prime$. $$\begin{eqnarray*} y'' &=&\dfrac{d}{dx}[ (x^{2}+2x+1) e^{x}] {=}(x^{2}+2x+1) \left( \dfrac{d}{dx}e^{x}\right) +\left[ \dfrac{d}{dx} ( x^{2}+2x+1) \right] e^{x} \\[-6.4pt] &&\hspace{4.8pc}\underset{\color{#0066A7}{\scriptsize \hbox{Product Rule}}}{\color{#0066A7}{\uparrow }}\\ &=&(x^{2}+2x+1) e^{x}+( 2x+2) e^{x}=(x^{2}+4x+3) e^{x}\\[5pt] y''' &=&\dfrac{d}{dx}[ ( x^{2}+4x+3) e^{x}] {=}(x^{2}+4x+3) \dfrac{d}{dx}e^{x}+\left[ \dfrac{d}{dx}(x^{2}+4x+3) \right] e^{x}\\[-6.4pt] &&\hspace{4.8pc}\underset{\color{#0066A7}{\scriptsize \hbox{Product Rule}}}{\color{#0066A7}{\uparrow }}\\ &=&( x^{2}+4x+3) e^{x}+\left( 2x+4\right) e^{x}=(x^{2}+6x+7) e^{x} \end{eqnarray*}$$

NOW WORK

Problem 45.

DEFINITION Acceleration

When the distance $s$ of an object from the origin at time $t$ is given by a function $s=s( t)$, the first derivative $\dfrac{ds}{dt}$ is the velocity $v=v( t)$ of the object at time $t$.

The acceleration $a=a( t)$ of an object at time $t$ is defined as the rate of change of velocity with respect to time. That is, $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt] a=a( t) =\dfrac{dv}{dt}=\dfrac{d}{dt}v=\dfrac{d}{dt} \! \left( \dfrac{ds}{dt}\right) =\dfrac{d^{2}s}{dt^{2}} }$$

IN WORDS

Acceleration is the second derivative of distance with respect to time.

A ball is propelled vertically upward from the ground with an initial velocity of 29.4 meters per second. The height $s$ (in meters) of the ball above the ground is approximately $s=s( t) =-4.9t^{2}+29.4t$, where $t$ is the number of seconds that elapse from the moment the ball is released.

Solution (a) Since $s=-4.9t^{2}+29.4t$, then $$\begin{eqnarray*} v&=&v( t) ={\dfrac{ds}{dt}}=-9.8t+29.4\\[5pt] v(1) &=&-9.8+29.4=19.6 \end{eqnarray*}$$

At $t=1$, the velocity of the ball is $19.6m/s$.

(b) The ball reaches its maximum height when $v( t) =0$. $$\begin{eqnarray*} v( t) &=&-9.8t+29.4=0 \\ 9.8t &=& 29.4 \\ t &=&3 \end{eqnarray*}$$

The ball reaches its maximum height after 3 seconds.

(c) The maximum height is $$s=s(3) =-4.9(3^{2})+29.4(3)=44.1$$

The maximum height of the ball is $44.1$ m.

(d) The acceleration of the ball at any time $t$ is $$a=a( t) ={\dfrac{d^{2}s}{dt^{2}}}=\dfrac{dv}{dt}=\dfrac{d}{dt}( -9.8t+29.4) =-9.8m/s^{2}$$

(e) There are two ways to answer the question “How long is the ball in the air?” First way: Since it takes $3$s for the ball to reach its maximum altitude, it follows that it will take another $3$s to reach the ground, for a total time of $6$s in the air. Second way: When the ball reaches the ground, $s=s( t) =0$. Solve for $t$: $$\begin{eqnarray*} s( t) &=&-4.9t^{2}+29.4t=0\\[4pt] && t (-4.9t + 29.4) =0\\[4pt] t&=&0\qquad \hbox{or}\qquad t=\frac{29.4}{4.9}=6 \end{eqnarray*}$$

Figure 25

The ball is at ground level at $t=0$ and at $t=6$, so the ball is in the air for 6 seconds.

(f) Upon impact with the ground, $t=6s$. So the velocity is $$v( 6) =(-9.8)(6)+29.4=-29.4$$

Upon impact the direction of the ball is downward, and its speed is 29.4m/s.

(g) The total distance traveled by the ball is $$s(3) +s(3) =2~s(3) =2( 44.1) =88.2{m}$$

See Figure 25 for an illustration.

NOW WORK

Problem 83.

NOTE

The Earth is not perfectly round; it bulges sightly at the equator, and its mass is not distributed uniformly. As a result, the acceleration of a freely falling body varies slightly.