Chapter 3.2, Problem 46E

(a)

To determine

To find: The derivative of $P\left(x\right)$ at $x=2$.

Answer to Problem 46E

The derivative of $P\left(x\right)$ at $x=2$ is $P^{\prime }\left(2\right)=\frac{3}{2}$.

Explanation of Solution

Given:

The function is $P\left(x\right)=F\left(x\right)G\left(x\right)$.

Derivative rule: Product Rule

If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then

$\frac{d}{dx}\left[f_1\left(x\right)f_2\left(x\right)\right]=f_1\left(x\right)\frac{d}{dx}\left[f_2\left(x\right)\right]+f_2\left(x\right)\frac{d}{dx}\left[f_1\left(x\right)\right]$

Calculation:

Obtain the derivative of $P\left(x\right)$ at $x=2$.

$P^{\prime }\left(x\right)=\frac{d}{dx}\left(F\left(x\right)G\left(x\right)\right)$

Apply the product rule (1) and simplify the terms,

$\begin{array}{c}{P}^{\prime }\left(x\right)=F\left(x\right)\frac{d}{dx}\left(G\left(x\right)\right)+G\left(x\right)\frac{d}{dx}\left(F\left(x\right)\right)\\ =F\left(x\right)G^{\prime }\left(x\right)+G\left(x\right)F^{\prime }\left(x\right)\end{array}$

Substitute 2 for x in $P^{\prime }\left(x\right)$,

$P^{\prime }\left(x\right)=F\left(x\right)G^{\prime }\left(x\right)+G\left(x\right)F^{\prime }\left(x\right)$ (1)

From the given graph, it is observed that $F\left(2\right)=3$ and $G\left(2\right)=2$, the function $F\left(x\right)$ is horizontal tangent at $x=2$ and the function $G\left(x\right)$ is linear that passes through the points $\left(0,1\right)\text{ and }\left(4,3\right)$.

Obtain the values $F^{\prime }\left(2\right)\text{ and }{G}^{\prime }\left(2\right)$.

The slope of tangent to $F\left(x\right)$ at $x=2$ is zero.

Therefore, the value $F^{\prime }\left(2\right)=0$.

The slope of the function $G\left(x\right)$ passing through the points $\left(0,1\right)\text{ and }\left(4,3\right)$ is,

$\begin{array}{c}{m}_1=\frac{3-1}{4-0}\\ =\frac{2}{4}\\ =\frac{1}{2}\end{array}$

Therefore, the value of $G^{\prime }\left(2\right)=\frac{1}{2}$.

Substitute the values $F\left(2\right)=3\text{ , }G\left(2\right)=2,F^{\prime }\left(2\right)=0\text{ and }{G}^{\prime }\left(2\right)=\frac{1}{2}$ in equation (1),

$\begin{array}{c}{P}^{\prime }\left(2\right)=\left(3\times \frac{1}{2}\right)+\left(2\times 0\right)\\ =\frac{3}{2}+0\\ =\frac{3}{2}\end{array}$

Therefore, the derivative of $P\left(x\right)$ at $x=2$ is $P^{\prime }\left(2\right)=\frac{3}{2}$.

(b)

To determine

To find: The derivative of $Q\left(x\right)$ at $x=7$.

Answer to Problem 46E

The derivative of $Q\left(x\right)$ at $x=7$ is $Q^{\prime }\left(7\right)=\frac{43}{12}$.

Explanation of Solution

Given:

The function is $Q\left(x\right)=\frac{F\left(x\right)}{G\left(x\right)}$.

Derivative rule: Quotient Rule

If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then

$\frac{d}{dx}\left[\frac{f_1\left(x\right)}{f_2\left(x\right)}\right]=\frac{f_2\left(x\right)\frac{d}{dx}\left[f_1\left(x\right)\right]-f_1\left(x\right)\frac{d}{dx}\left[f_2\left(x\right)\right]}{{\left[f_{2}x\right]}^2}$

Calculation:

Obtain the derivative of $Q\left(x\right)$ at $x=7$.

$Q^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{F\left(x\right)}{G\left(x\right)}\right)$

Apply the Quotient rule (1) and simplify the terms,

$\begin{array}{c}{Q}^{\prime }\left(x\right)=\frac{G\left(x\right)\frac{d}{dx}\left(F\left(x\right)\right)-F\left(x\right)\frac{d}{dx}\left(G\left(x\right)\right)}{{\left(G\left(x\right)\right)}^2}\\ =\frac{G\left(x\right)F^{\prime }\left(x\right)-F\left(x\right)G^{\prime }\left(x\right)}{{\left(G\left(x\right)\right)}^2}\end{array}$

Substitute 7 for x in $Q^{\prime }\left(x\right)$,

$Q^{\prime }\left(7\right)=\frac{G\left(7\right)F^{\prime }\left(7\right)-F\left(7\right)G^{\prime }\left(7\right)}{{\left(G\left(7\right)\right)}^2}$ (2)

From the given graph, it is observed that $F\left(7\right)=5\text{ , }G\left(7\right)=1$ , the function $F\left(x\right)$ is linear that passes through the points $\left(3,4\right)\text{ and }\left(7,5\right)$ and the function $G\left(x\right)$ is linear that passes through the points $\left(4,3\right)\text{ and }\left(7,1\right)$.

Obtain the values $F^{\prime }\left(7\right)\text{ and }{G}^{\prime }\left(5\right)$.

Since the slope of the line at every point is equal, the slope of the function $f\left(x\right)$ passing through the points $\left(3,4\right)\text{ and }\left(7,5\right)$ is,

$\begin{array}{c}{m}_2=\frac{5-4}{7-3}\\ =\frac{1}{4}\end{array}$

Therefore, the value of $F^{\prime }\left(7\right)=\frac{1}{4}$.

The slope of the function $G\left(x\right)$ passing through the points $\left(4,3\right)\text{ and }\left(7,1\right)$ is,

$\begin{array}{c}{m}_3=\frac{1-3}{7-4}\\ =\frac{-2}{3}\end{array}$

Therefore, the value of $G^{\prime }\left(7\right)=\frac{-2}{3}$.

Substitute the values $F\left(7\right)=5\text{ , G}\left(7\right)=1,F^{\prime }\left(7\right)=\frac{1}{4}\text{ and }{G}^{\prime }\left(7\right)=\frac{-2}{3}$ in equation (2),

$\begin{array}{c}{Q}^{\prime }\left(7\right)=\frac{\left(1\times \frac{1}{4}\right)-\left(5\times \frac{-2}{3}\right)}{1^2}\\ =\frac{1}{4}+\frac{10}{3}\\ =\frac{3}{12}+\frac{40}{12}\\ =\frac{43}{12}\end{array}$

Therefore, the derivative of $Q\left(x\right)$ at $x=7$ is $Q^{\prime }\left(7\right)=\frac{43}{12}$.