Chapter 3.1, Problem 62E

(a)

To determine

To find : derivative of $2f\left(x\right)$ with respect to $x$ at $x=2$ .

Answer to Problem 62E

Derivative of $2f\left(x\right)$ with respect to $x$ at $x=2$ is $\frac{2}{3}$ .

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=2$ and, $x=3$ .

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 2& 8& 2& 1/3& -3\\ 3& 3& -4& 2\pi & 5\end{array}$

Formula needed :

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Differentiate $2f\left(x\right)$ with respect to $x$ .

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

Substitute $x=2$ in the $\frac{d}{dx}2f\left(x\right)=2f^{\prime }\left(x\right)$ .

  $\begin{array}{c}\frac{d}{dx}2f\left(2\right)=2f^{\prime }\left(2\right)\\ =2\cdot \frac{1}{3}\\ =2/3\end{array}$

Hence, derivative of $2f\left(x\right)$ with respect to $x$ at $x=2$ is $\frac{2}{3}$ .

(b)

To determine

To find : derivative of $f\left(x\right)+g\left(x\right)$ with respect to $x$ at $x=3$ .

Answer to Problem 62E

Derivative of $f\left(x\right)+g\left(x\right)$ with respect to $x$ at $x=3$ is $2\pi +5$ .

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=2$ and, $x=3$ .

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 2& 8& 2& 1/3& -3\\ 3& 3& -4& 2\pi & 5\end{array}$

Formula needed :

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Differentiate $f\left(x\right)+g\left(x\right)$ with respect to $x$ .

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

Substitute $x=3$ in the $\frac{d}{dx}\left(f\left(x\right)+g\left(x\right)\right)=f^{\prime }\left(x\right)+g^{\prime }\left(x\right)$ .

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

Hence, derivative of $f\left(x\right)+g\left(x\right)$ with respect to $x$ at $x=3$ is $2\pi +5$ .

(c)

To determine

To find : derivative of $f\left(x\right)\cdot g\left(x\right)$ with respect to $x$ at $x=3$ .

Answer to Problem 62E

Derivative of $f\left(x\right)\cdot g\left(x\right)$ with respect to $x$ at $x=3$ is $-8\pi +15$ .

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=2$ and, $x=3$ .

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 2& 8& 2& 1/3& -3\\ 3& 3& -4& 2\pi & 5\end{array}$

Formula needed :

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Differentiate $f\left(x\right)\cdot g\left(x\right)$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\cdot g\left(x\right)\right)=g\left(x\right)\cdot \frac{d}{dx}f\left(x\right)+f\left(x\right)\cdot \frac{d}{dx}g\left(x\right)\\ =g\left(x\right)\cdot f^{\prime }\left(x\right)+f\left(x\right)\cdot g^{\prime }\left(x\right)\end{array}$

Substitute $x=3$ in the $\frac{d}{dx}\left(f\left(x\right)\cdot g\left(x\right)\right)=g\left(x\right)\cdot f^{\prime }\left(x\right)+f\left(x\right)\cdot g^{\prime }\left(x\right)$ .

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

Hence, derivative of $f\left(x\right)\cdot g\left(x\right)$ with respect to $x$ at $x=3$ is $-8\pi +15$ .

(d)

To determine

To find : derivative of $f\left(x\right)/g\left(x\right)$ with respect to $x$ at $x=2$ .

Answer to Problem 62E

Derivative of $f\left(x\right)/g\left(x\right)$ with respect to $x$ at $x=2$ is $\frac{-35}{6}$ .

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=2$ and, $x=3$ .

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 2& 8& 2& 1/3& -3\\ 3& 3& -4& 2\pi & 5\end{array}$

Formula needed :

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Quotient rule of derivative: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }\cdot v-u\cdot v^{\prime }}{v^2}$

Differentiate $f\left(x\right)/g\left(x\right)$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)/g\left(x\right)\right)=\frac{g\left(x\right)\cdot \frac{d}{dx}f\left(x\right)-f\left(x\right)\cdot \frac{d}{dx}g\left(x\right)}{{\left(g\left(x\right)\right)}^2}\\ =\frac{g\left(x\right)\cdot f^{\prime }\left(x\right)+f\left(x\right)\cdot g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}\end{array}$

Substitute $x=2$ in the $\frac{d}{dx}\left(f\left(x\right)/g\left(x\right)\right)=\frac{g\left(x\right)\cdot f^{\prime }\left(x\right)+f\left(x\right)\cdot g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}$ .

  $\begin{array}{c}\frac{d}{dx}\left(f\left(2\right)/g\left(2\right)\right)=\frac{g\left(2\right)\cdot f^{\prime }\left(2\right)+f\left(2\right)\cdot g^{\prime }\left(2\right)}{{\left(g\left(2\right)\right)}^2}\\ =\frac{2\cdot 1/3+8\cdot \left(-3\right)}{{\left(2\right)}^2}\\ =\frac{2/3-24}{4}\\ =\frac{-35}{6}\end{array}$

Hence, derivative of $f\left(x\right)/g\left(x\right)$ with respect to $x$ at $x=2$ is $\frac{-35}{6}$ .

(e)

To determine

To find : derivative of $f\left(g\left(x\right)\right)$ with respect to $x$ at $x=2$ .

Answer to Problem 62E

Derivative of $f\left(g\left(x\right)\right)$ with respect to $x$ at $x=2$ is $-1$ .

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=2$ and, $x=3$ .

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 2& 8& 2& 1/3& -3\\ 3& 3& -4& 2\pi & 5\end{array}$

Formula needed :

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Quotient rule of derivative: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }\cdot v-u\cdot v^{\prime }}{v^2}$

Chain rule of derivative: $\frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$

Differentiate $f\left(g\left(x\right)\right)$ with respect to $x$ .

  $\frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$

Substitute $x=2$ in the $\frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$ .

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

Hence, derivative of $f\left(g\left(x\right)\right)$ with respect to $x$ at $x=2$ is $-1$ .

(f)

To determine

To find : derivative of $\sqrt{f\left(x\right)}$ with respect to $x$ at $x=2$ .

Answer to Problem 62E

Derivative of $\sqrt{f\left(x\right)}$ with respect to $x$ at $x=2$ is $\frac{1}{12\sqrt{2}}$ .

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=2$ and, $x=3$ .

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 2& 8& 2& 1/3& -3\\ 3& 3& -4& 2\pi & 5\end{array}$

Formula needed :

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Quotient rule of derivative: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }\cdot v-u\cdot v^{\prime }}{v^2}$

Chain rule of derivative: $\frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$

Differentiate $\sqrt{f\left(x\right)}$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\sqrt{f\left(x\right)}=\frac{1}{2\sqrt{f\left(x\right)}}\frac{d}{dx}f\left(x\right)\\ =\frac{1}{2\sqrt{f\left(x\right)}}{f}^{\prime }\left(x\right)\\ =\frac{f^{\prime }\left(x\right)}{2\sqrt{f\left(x\right)}}\end{array}$

Substitute $x=2$ in the $\frac{d}{dx}\sqrt{f\left(x\right)}=\frac{f^{\prime }\left(x\right)}{2\sqrt{f\left(x\right)}}$ .

  $\begin{array}{c}\frac{d}{dx}\sqrt{f\left(2\right)}=\frac{f^{\prime }\left(2\right)}{2\sqrt{f\left(2\right)}}\\ =\frac{1/3}{2\sqrt{8}}\\ =\frac{1}{12\sqrt{2}}\end{array}$

Hence, derivative of $\sqrt{f\left(x\right)}$ with respect to $x$ at $x=2$ is $\frac{1}{12\sqrt{2}}$ .

(g)

To determine

To find : derivative of $1/g^2\left(x\right)$ with respect to $x$ at $x=3$ .

Answer to Problem 62E

Derivative of $1/g^2\left(x\right)$ with respect to $x$ at $x=3$ is $\frac{5}{32}$ .

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=2$ and, $x=3$ .

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 2& 8& 2& 1/3& -3\\ 3& 3& -4& 2\pi & 5\end{array}$

Formula needed :

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Quotient rule of derivative: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }\cdot v-u\cdot v^{\prime }}{v^2}$

Chain rule of derivative: $\frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$

Differentiate $1/g^2\left(x\right)$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}1/g^2\left(x\right)=\frac{-2}{g^3\left(x\right)}\frac{d}{dx}g\left(x\right)\\ =-\frac{2}{g^3\left(x\right)}{g}^{\prime }\left(x\right)\\ =-\frac{2g^{\prime }\left(x\right)}{g^3\left(x\right)}\end{array}$

Substitute $x=3$ in the $\frac{d}{dx}1/g^2\left(x\right)=-\frac{2g^{\prime }\left(x\right)}{g^3\left(x\right)}$ .

  $\begin{array}{c}\frac{d}{dx}1/g^2\left(3\right)=-\frac{2g^{\prime }\left(3\right)}{g^3\left(3\right)}\\ =-\frac{2\cdot 5}{{\left(-4\right)}^3}\\ =-\frac{10}{-64}\\ =\frac{5}{32}\end{array}$

Hence, derivative of $1/g^2\left(x\right)$ with respect to $x$ at $x=3$ is $\frac{5}{32}$ .

(h)

To determine

To find : derivative of $\sqrt{f^2\left(x\right)+g^2\left(x\right)}$ with respect to $x$ at $x=2$ .

Answer to Problem 62E

Derivative of $\sqrt{f^2\left(x\right)+g^2\left(x\right)}$ with respect to $x$ at $x=2$ is $\frac{-10}{3\sqrt{68}}$ .

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=2$ and, $x=3$ .

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 2& 8& 2& 1/3& -3\\ 3& 3& -4& 2\pi & 5\end{array}$

Formula needed :

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Quotient rule of derivative: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }\cdot v-u\cdot v^{\prime }}{v^2}$

Chain rule of derivative: $\frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$

Differentiate $\sqrt{f^2\left(x\right)+g^2\left(x\right)}$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\sqrt{f^2\left(x\right)+g^2\left(x\right)}=\frac{1}{2\sqrt{f^2\left(x\right)+g^2\left(x\right)}}\frac{d}{dx}\left(f^2\left(x\right)+g^2\left(x\right)\right)\\ =\frac{1}{2\sqrt{f^2\left(x\right)+g^2\left(x\right)}}\left(2f\left(x\right)\cdot f^{\prime }\left(x\right)+2g\left(x\right)\cdot g^{\prime }\left(x\right)\right)\\ =\frac{2f\left(x\right)\cdot f^{\prime }\left(x\right)+2g\left(x\right)\cdot g^{\prime }\left(x\right)}{2\sqrt{f^2\left(x\right)+g^2\left(x\right)}}\\ =\frac{f\left(x\right)\cdot f^{\prime }\left(x\right)+g\left(x\right)\cdot g^{\prime }\left(x\right)}{\sqrt{f^2\left(x\right)+g^2\left(x\right)}}\end{array}$

Substitute $x=2$ in the $\frac{d}{dx}\sqrt{f^2\left(x\right)+g^2\left(x\right)}=\frac{f\left(x\right)\cdot f^{\prime }\left(x\right)+g\left(x\right)\cdot g^{\prime }\left(x\right)}{\sqrt{f^2\left(x\right)+g^2\left(x\right)}}$ .

  $\begin{array}{c}\frac{d}{dx}\sqrt{f^2\left(2\right)+g^2\left(2\right)}=\frac{f\left(2\right)\cdot f^{\prime }\left(2\right)+g\left(2\right)\cdot g^{\prime }\left(2\right)}{\sqrt{f^2\left(2\right)+g^2\left(2\right)}}\\ =\frac{8\cdot 1/3+2\cdot \left(-3\right)}{\sqrt{{\left(8\right)}^2+{\left(2\right)}^2}}\\ =\frac{8/3-6}{\sqrt{64+4}}\\ =\frac{-10}{3\sqrt{68}}\end{array}$

Hence, derivative of $\sqrt{f^2\left(x\right)+g^2\left(x\right)}$ with respect to $x$ at $x=2$ is $\frac{-10}{3\sqrt{68}}$ .