Chapter 3.1, Problem 64E

(a)

To determine

To find : derivative of $5f\left(x\right)-g\left(x\right)$ with respect to $x$ at $x=1$.

Answer to Problem 64E

Derivative of $5f\left(x\right)-g\left(x\right)$ with respect to $x$ at $x=1$ is $1$.

Explanation of Solution

Given information :

The function

  $f\left(x\right),g\left(x\right)$ and their derivatives at $x=0$ and, $x=1$.

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 0& 1& 1& 5& 1/3\\ 1& 3& -4& -1/3& -8/3\end{array}$

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 $5f\left(x\right)-g\left(x\right)$ with respect to $x$.

  $\begin{array}{c}\frac{d}{dx}\left(5f\left(x\right)-g\left(x\right)\right)=5\frac{d}{dx}f\left(x\right)-\frac{d}{dx}g\left(x\right)\\ =5f^{\prime }\left(x\right)-g^{\prime }\left(x\right)\end{array}$

Substitute $x=1$ in the $\frac{d}{dx}\left(5f\left(x\right)-g\left(x\right)\right)=5f^{\prime }\left(x\right)-g^{\prime }\left(x\right)$.

  $\begin{array}{c}\frac{d}{dx}\left(5f\left(1\right)-g\left(1\right)\right)=5f^{\prime }\left(1\right)-g^{\prime }\left(1\right)\\ =5\left(-1/3\right)-\left(-8/3\right)\\ =-\frac{5}{3}+\frac{8}{3}\\ =1\end{array}$

Hence, derivative of $5f\left(x\right)-g\left(x\right)$ with respect to $x$ at $x=1$ is $1$.

(b)

To determine

To find : derivative of $f\left(x\right)g^3\left(x\right)$ with respect to $x$ at $x=0$.

Answer to Problem 64E

Derivative of $f\left(x\right)g^3\left(x\right)$ with respect to $x$ at $x=0$ is $6$.

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=0$ and, $x=1$.

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 0& 1& 1& 5& 1/3\\ 1& 3& -4& -1/3& -8/3\end{array}$

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 $f\left(x\right)g^3\left(x\right)$ with respect to $x$.

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

Substitute $x=0$ in the

  $\frac{d}{dx}\left(f\left(x\right)g^3\left(x\right)\right)=g^3\left(x\right)\cdot f^{\prime }\left(x\right)+f\left(x\right)\cdot 3g^2\left(x\right)g^{\prime }\left(x\right)$.

  $\begin{array}{c}\frac{d}{dx}\left(f\left(0\right)g^3\left(0\right)\right)=g^3\left(0\right)\cdot f^{\prime }\left(0\right)+f\left(0\right)\cdot 3g^2\left(0\right)g^{\prime }\left(0\right)\\ =1\cdot 5+1\cdot 3\cdot 1/3\\ =5+1\\ =6\end{array}$

Hence, derivative of $f\left(x\right)g^3\left(x\right)$ with respect to $x$ at $x=0$ is $6$.

(c)

To determine

To find : derivative of $\frac{f\left(x\right)}{g\left(x\right)+1}$ with respect to $x$ at $x=1$.

Answer to Problem 64E

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

Explanation of Solution

Given information :

The function

  $f\left(x\right),g\left(x\right)$ and their derivatives at $x=0$ and, $x=1$.

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 0& 1& 1& 5& 1/3\\ 1& 3& -4& -1/3& -8/3\end{array}$

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

  $\frac{f\left(x\right)}{g\left(x\right)+1}$ with respect to $x$.

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

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

  $\begin{array}{c}\frac{d}{dx}\left(\frac{f\left(1\right)}{g\left(1\right)+1}\right)=\frac{\left(g\left(1\right)+1\right)\cdot f^{\prime }\left(1\right)-f\left(1\right)\cdot g^{\prime }\left(1\right)}{{\left(g\left(1\right)+1\right)}^2}\\ =\frac{\left(-4+1\right)\cdot \left(-1/3\right)-3\cdot \left(-8/3\right)}{{\left(-4+1\right)}^2}\\ =\frac{-3\cdot \left(-1/3\right)+8}{{\left(-3\right)}^2}\\ =\frac{1+8}{9}\\ =1\end{array}$

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

(d)

To determine

To find : derivative of $f\left(g\left(x\right)\right)$ with respect to $x$ at $x=0$.

Answer to Problem 64E

Derivative of $f\left(g\left(x\right)\right)$ with respect to $x$ at $x=0$ is $5/3$.

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=0$ and, $x=1$.

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 0& 1& 1& 5& 1/3\\ 1& 3& -4& -1/3& -8/3\end{array}$

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 $x=0$ in the $\frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$.

  $\begin{array}{c}\frac{d}{dx}f\left(g\left(0\right)\right)=f^{\prime }\left(g\left(0\right)\right)\cdot g^{\prime }\left(0\right)\\ =f^{\prime }\left(1\right)\cdot \left(1/3\right)\\ =5\cdot \left(1/3\right)\\ =5/3\end{array}$

Hence, derivative of $f\left(g\left(x\right)\right)$ with respect to $x$ at $x=0$ is $5/3$.

(e)

To determine

To find : derivative of $g\left(f\left(x\right)\right)$ with respect to $x$ at $x=0$.

Answer to Problem 64E

Derivative of $g\left(f\left(x\right)\right)$ with respect to $x$ at $x=0$ is $-40/3$.

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=0$ and, $x=1$.

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 0& 1& 1& 5& 1/3\\ 1& 3& -4& -1/3& -8/3\end{array}$

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

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

Substitute $x=0$ in the $\frac{d}{dx}g\left(f\left(x\right)\right)=g^{\prime }\left(f\left(x\right)\right)\cdot f^{\prime }\left(x\right)$.

  $\begin{array}{c}\frac{d}{dx}g\left(f\left(0\right)\right)=g^{\prime }\left(f\left(0\right)\right)\cdot f^{\prime }\left(0\right)\\ =g^{\prime }\left(1\right)\cdot 5\\ =-8/3\cdot 5\\ =-40/3\end{array}$

Hence, derivative of $g\left(f\left(x\right)\right)$ with respect to $x$ at $x=0$ is $-40/3$.

(f)

To determine

To find : derivative of

  ${\left(g\left(x\right)+f\left(x\right)\right)}^{-2}$ with respect to $x$ at $x=1$.

Answer to Problem 64E

Derivative of

  ${\left(g\left(x\right)+f\left(x\right)\right)}^{-2}$ with respect to $x$ at $x=1$ is $-6$.

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=0$ and, $x=1$.

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 0& 1& 1& 5& 1/3\\ 1& 3& -4& -1/3& -8/3\end{array}$

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

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

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

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

Hence, derivative of ${\left(g\left(x\right)+f\left(x\right)\right)}^{-2}$ with respect to $x$ at $x=1$ is $-6$.

(g)

To determine

To find : derivative of $f\left(x+g\left(x\right)\right)$ with respect to $x$ at $x=0$.

Answer to Problem 64E

Derivative of $f\left(x+g\left(x\right)\right)$ with respect to $x$ at $x=0$ is $-4/9$.

Explanation of Solution

Given information :

The function $f\left(x\right),g\left(x\right)$ and their derivatives at $x=0$ and, $x=1$.

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ 0& 1& 1& 5& 1/3\\ 1& 3& -4& -1/3& -8/3\end{array}$

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

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

Substitute $x=0$ in the $\frac{d}{dx}f\left(x+g\left(x\right)\right)=f^{\prime }\left(x+g\left(x\right)\right)\cdot \left(1+g^{\prime }\left(x\right)\right)$.

  $\begin{array}{c}\frac{d}{dx}f\left(0+g\left(0\right)\right)=f^{\prime }\left(0+g\left(0\right)\right)\cdot \left(1+g^{\prime }\left(0\right)\right)\\ =f^{\prime }\left(1\right)\cdot \left(1+1/3\right)\\ =-1/3\cdot \left(4/3\right)\\ =-4/9\end{array}$

Hence, derivative of $f\left(x+g\left(x\right)\right)$ with respect to $x$ at $x=0$ is $-4/9$.