Chapter 4.1, Problem 56E

a.

To determine

To find: To find the first derivative of the function when $x=2\text{ and }x=3$ .

Answer to Problem 56E

The first derivative of $2f\left(x\right),\text{when }x=2$ is $\frac{2}{3}$ .

Explanation of Solution

Given: $2f\left(x\right),\text{when }x=2$

$x$	$f\left(x\right)$	$g\left(x\right)$	$f^{\prime }\left(x\right)$	$g^{\prime }\left(x\right)$
$2$	$8$	$2$	$\frac{1}{3}$	$-3$
$3$	$3$	$-4$	$2\pi$	$5$

Calculation:

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{dy}{dx}\left(2f\left(x\right)\right)\\ f^{\prime }\left(x\right)=2\left(f^{\prime }\left(x\right)\right)\\ f^{\prime }\left(2\right)=2\left(\frac{1}{3}\right)\\ f^{\prime }\left(2\right)=\frac{2}{3}\end{array}$

The first derivative of $2f\left(x\right),\text{when }x=2$ is $\frac{2}{3}$ .

b.

To determine

To find: To find the first derivative of the function when $x=2\text{ and }x=3$ .

Answer to Problem 56E

The first derivative of $f\left(x\right)+g\left(x\right),\text{when }x=3$ is $2\pi +5$ .

Explanation of Solution

Given: $f\left(x\right)+g\left(x\right),\text{when }x=3$

$x$	$f\left(x\right)$	$g\left(x\right)$	$f^{\prime }\left(x\right)$	$g^{\prime }\left(x\right)$
$2$	$8$	$2$	$\frac{1}{3}$	$-3$
$3$	$3$	$-4$	$2\pi$	$5$

Calculation:

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{dy}{dx}\left(f\left(x\right)+g\left(x\right)\right)\\ f^{\prime }\left(x\right)=\left(f^{\prime }\left(x\right)+g^{\prime }\left(x\right)\right)\\ f^{\prime }\left(3\right)=\left(f^{\prime }\left(3\right)+g^{\prime }\left(3\right)\right)\\ f^{\prime }\left(3\right)=2\pi +5\end{array}$

The first derivative of $f\left(x\right)+g\left(x\right),\text{when }x=3$ is $2\pi +5$ .

c.

To determine

To find: To find the first derivative of the function when $x=2\text{ and }x=3$ .

Answer to Problem 56E

The first derivative of $f\left(x\right)\cdot g\left(x\right),\text{when }x=3$ is $-8\pi +15$ .

Explanation of Solution

Given: $f\left(x\right)\cdot g\left(x\right),\text{when }x=3$

$x$	$f\left(x\right)$	$g\left(x\right)$	$f^{\prime }\left(x\right)$	$g^{\prime }\left(x\right)$
$2$	$8$	$2$	$\frac{1}{3}$	$-3$
$3$	$3$	$-4$	$2\pi$	$5$

Calculation:

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{dy}{dx}\left(f\left(x\right)\cdot g\left(x\right)\right)\\ f^{\prime }\left(x\right)=f^{\prime }\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot g^{\prime }\left(x\right)\\ f^{\prime }\left(3\right)=f^{\prime }\left(3\right)\cdot g\left(3\right)+f\left(3\right)\cdot g^{\prime }\left(3\right)\\ f^{\prime }\left(3\right)=2\pi \times \left(-4\right)+3\times 5\\ f^{\prime }\left(3\right)=-8\pi +15\end{array}$

The first derivative of $f\left(x\right)\cdot g\left(x\right),\text{when }x=3$ is $-8\pi +15$ .

d.

To determine

To find: To find the first derivative of the function when $x=2\text{ and }x=3$ .

Answer to Problem 56E

The first derivative of $\frac{f\left(x\right)}{g\left(x\right)},\text{when }x=2$ is $\frac{37}{6}$ .

Explanation of Solution

Given: $\frac{f\left(x\right)}{g\left(x\right)},\text{when }x=2$

$x$	$f\left(x\right)$	$g\left(x\right)$	$f^{\prime }\left(x\right)$	$g^{\prime }\left(x\right)$
$2$	$8$	$2$	$\frac{1}{3}$	$-3$
$3$	$3$	$-4$	$2\pi$	$5$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{dy}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)\\ f^{\prime }\left(x\right)=\frac{f^{\prime }\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}\\ f^{\prime }\left(2\right)=\frac{f^{\prime }\left(2\right)\cdot g\left(2\right)-f\left(2\right)\cdot g^{\prime }\left(2\right)}{{\left(g\left(2\right)\right)}^2}\\ f^{\prime }\left(2\right)=\frac{\frac{1}{3}\cdot 2-8\cdot \left(-3\right)}{{\left(2\right)}^2}\\ f^{\prime }\left(2\right)=\frac{\frac{2}{3}+24}{4}\\ f^{\prime }\left(2\right)=\frac{74}{12}\\ f^{\prime }\left(2\right)=\frac{37}{6}\end{array}$

The first derivative of $\frac{f\left(x\right)}{g\left(x\right)},\text{when }x=2$ is $\frac{37}{6}$ .

e.

To determine

To find: To find the first derivative of the function when $x=2\text{ and }x=3$ .

Answer to Problem 56E

The first derivative of $f\left(g\left(x\right)\right),\text{when }x=2$ is $-1$ .

Explanation of Solution

Given: $2f\left(x\right),\text{when }x=2$

$x$	$f\left(x\right)$	$g\left(x\right)$	$f^{\prime }\left(x\right)$	$g^{\prime }\left(x\right)$
$2$	$8$	$2$	$\frac{1}{3}$	$-3$
$3$	$3$	$-4$	$2\pi$	$5$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{dy}{dx}f\left(g\left(x\right)\right)\\ f^{\prime }\left(x\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)\\ f^{\prime }\left(2\right)=f^{\prime }\left(g\left(2\right)\right)\cdot g^{\prime }\left(2\right)\\ f^{\prime }\left(2\right)=f^{\prime }\left(2\right)\cdot g^{\prime }\left(2\right)\\ f^{\prime }\left(2\right)=\frac{1}{3}\left(-3\right)\\ f^{\prime }\left(2\right)=-1\end{array}$

The first derivative of $f\left(g\left(x\right)\right),\text{when }x=2$ is $-1$ .

f.

To determine

To find: To find the first derivative of the function when $x=2\text{ and }x=3$ .

Answer to Problem 56E

The first derivative of $\sqrt{f\left(x\right)},\text{when }x=2$ is $\frac{1}{12\sqrt{2}}$ .

Explanation of Solution

Given: $\sqrt{f\left(x\right)},\text{when }x=2$

$x$	$f\left(x\right)$	$g\left(x\right)$	$f^{\prime }\left(x\right)$	$g^{\prime }\left(x\right)$
$2$	$8$	$2$	$\frac{1}{3}$	$-3$
$3$	$3$	$-4$	$2\pi$	$5$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{dy}{dx}\sqrt{f\left(x\right)}\\ f^{\prime }\left(x\right)=\frac{1}{2}f{\left(x\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cdot f^{\prime }\left(x\right)\\ f^{\prime }\left(2\right)=\frac{1}{2}{\left(8\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \frac{1}{3}\\ f^{\prime }\left(2\right)=\frac{1}{12\sqrt{2}}\end{array}$

The first derivative of $\sqrt{f\left(x\right)},\text{when }x=2$ is $\frac{1}{12\sqrt{2}}$ .

g.

To determine

To find: To find the first derivative of the function when $x=2\text{ and }x=3$ .

Answer to Problem 56E

The first derivative of $\frac{1}{g^2\left(x\right)},\text{when }x=3$ is $\frac{5}{32}$ .

Explanation of Solution

Given: $\frac{1}{g^2\left(x\right)},\text{when }x=3$

$x$	$f\left(x\right)$	$g\left(x\right)$	$f^{\prime }\left(x\right)$	$g^{\prime }\left(x\right)$
$2$	$8$	$2$	$\frac{1}{3}$	$-3$
$3$	$3$	$-4$	$2\pi$	$5$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{dy}{dx}\left(\frac{1}{g^2\left(x\right)}\right)\\ f^{\prime }\left(x\right)=\frac{dy}{dx}{\left(g^2\left(x\right)\right)}^{-2}\\ f^{\prime }\left(3\right)=-2g{\left(3\right)}^{-2}\cdot g^{\prime }\left(x\right)\\ f^{\prime }\left(3\right)=-2\times \frac{1}{{\left(-4\right)}^3}\times 5\\ f^{\prime }\left(3\right)=\frac{5}{32}\end{array}$

The first derivative of $\frac{1}{g^2\left(x\right)},\text{when }x=3$ is $\frac{5}{32}$ .

h.

To determine

To find: To find the first derivative of the function when $x=2\text{ and }x=3$ .

Answer to Problem 56E

The first derivative of $\sqrt{f^2\left(x\right)+g^2\left(x\right)},\text{when }x=2$ is $\frac{-5}{3\sqrt{17}}$ .

Explanation of Solution

Given: $\sqrt{f^2\left(x\right)+g^2\left(x\right)},\text{when }x=2$

$x$	$f\left(x\right)$	$g\left(x\right)$	$f^{\prime }\left(x\right)$	$g^{\prime }\left(x\right)$
$2$	$8$	$2$	$\frac{1}{3}$	$-3$
$3$	$3$	$-4$	$2\pi$	$5$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{dy}{dx}\left({\left(f^2\left(x\right)+g^2\left(x\right)\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\\ f^{\prime }\left(x\right)=\frac{1}{2}{\left(f^2\left(x\right)+g^2\left(x\right)\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ f^{\prime }\left(x\right)=\frac{1}{2{\left(f^2\left(x\right)+g^2\left(x\right)\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ f^{\prime }\left(x\right)=\frac{1}{2\sqrt{\left(f^2\left(x\right)+g^2\left(x\right)\right)}}\times \left(2f\left(x\right)f^{\prime }\left(x\right)+2g\left(x\right)g^{\prime }\left(x\right)\right)\\ f^{\prime }\left(2\right)=\frac{1}{2\sqrt{\left(f^2\left(2\right)+g^2\left(2\right)\right)}}\times \left(2f\left(2\right)f^{\prime }\left(2\right)+2g\left(2\right)g^{\prime }\left(2\right)\right)\\ f^{\prime }\left(2\right)=\frac{-5}{3\sqrt{17}}\end{array}$

The first derivative of $\sqrt{f^2\left(x\right)+g^2\left(x\right)},\text{when }x=2$ is $\frac{-5}{3\sqrt{17}}$ .