Chapter 2, Problem 68RE

(a).

To determine

To find: The first derivative of the function at a given point.

Answer to Problem 68RE

First derivative of $3f(x)-g(x)$ at $x=-1$ is 5.

Explanation of Solution

Given:

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ -1& 0& -1& 2& 1\\ 0& -1& -3& -2& 4\end{array}$

Concept Used:

  $\frac{d}{dx}\left[cf(x)\right]=c\cdot \frac{d}{dx}\left[f(x)\right]$

  $\frac{d}{dx}\left[f(x)-g(x)\right]=\frac{d}{dx}\left[f(x)\right]-\frac{d}{dx}\left[g\left(x\right)\right]$

Calculation:

  $\begin{array}{c}\frac{d}{dx}\left[3\cdot f\left(x\right)-g\left(x\right)\right]=3\cdot \frac{d}{dx}\left[f\left(x\right)\right]-\frac{d}{dx}\left[g\left(x\right)\right]\\ =3\cdot f\text{'}(x)-g\text{'}\left(x\right)\end{array}$

From the given table:

  $\begin{array}{c}\text{At}x=-1,\\ {\left[\frac{d}{dx}\left[3\cdot f\left(x\right)-g\left(x\right)\right]\right]}_{x=-1}=3\cdot f\text{'}(-1)-g\text{'}\left(-1\right)\\ =3\times 2-1\\ =5\end{array}$

b.

To determine

To calculate: The first derivative of the function at a given point.

Answer to Problem 68RE

Derivative is 2.

Explanation of Solution

Given:

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ -1& 0& -1& 2& 1\\ 0& -1& -3& -2& 4\end{array}$

Concept Used:

  $\frac{d}{dx}\left[u\cdot v\right]=u\frac{d}{dx}\left[v\right]+v\frac{d}{dx}\left[u\right]$ .

Calculation:

  $\begin{array}{c}\frac{d}{dx}\left[f\left(x\right)\cdot g\left(x\right)\right]=f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]+g\left(x\right)\frac{d}{dx}\left[f(x)\right]\\ =f\left(x\right)\cdot g\text{'}\left(x\right)+g\left(x\right)\cdot f\text{'}\left(x\right)\\ \text{At}x=0,\\ {\left[\frac{d}{dx}\left[f\left(x\right)\cdot g(x)\right]\right]}_{x=0}=f\left(0\right)\cdot g\text{'}\left(0\right)+g\left(0\right)\cdot f\text{'}\left(0\right)\\ =-1\times 4+\left(-3\right)\left(-2\right)\\ =-4+6\\ =2\end{array}$

Conclusion:

Derivative is 2.

c.

To determine

To calculate: The first derivative of the function at a given point.

Answer to Problem 68RE

The derivative is 0.

Explanation of Solution

Given:

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ -1& 0& -1& 2& 1\\ 0& -1& -3& -2& 4\end{array}$

Concept Used:

  $\frac{d}{dx}\left[u\cdot v\right]=u\frac{d}{dx}\left[v\right]+v\frac{d}{dx}\left[u\right]$ .

Calculation:

  $\begin{array}{c}\frac{d}{dx}\left[f\left(x\right)\cdot g\left(x\right)\right]=f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]+g\left(x\right)\frac{d}{dx}\left[f(x)\right]\\ =f\left(x\right)\cdot g\text{'}\left(x\right)+g\left(x\right)\cdot f\text{'}\left(x\right)\\ Atx=-1,\\ {\left[\frac{d}{dx}\left[f\left(x\right)\cdot g(x)\right]\right]}_{x=0}=f\left(-1\right)\cdot g\text{'}\left(-1\right)+g\left(-1\right)\cdot f\text{'}\left(-1\right)\\ =0\left(1\right)+1\left(0\right)\\ =0\end{array}$

Conclusion:

Derivative is 0

d.

To determine

To calculate: The first derivative of the function at a given point.

Answer to Problem 68RE

The derivative is $\frac{10}{9}$ .

Explanation of Solution

Given:

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ -1& 0& -1& 2& 1\\ 0& -1& -3& -2& 4\end{array}$

Concept Used:

  $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{\frac{d}{dx}\left[u\left(x\right)\right]\cdot v(x)-u(x)\cdot \frac{d}{dx}\left[v(x)\right]}{v{(x)}^2}$

Calculation:

  $\begin{array}{c}\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{\frac{d}{dx}\left[f\left(x\right)\right]\cdot g(x)-f(x)\cdot \frac{d}{dx}\left[g(x)\right]}{{\left[g(x)\right]}^2}\\ =\frac{g(x)\cdot f\text{'}\left(x\right)-f\left(x\right)\cdot g\text{'}(x)}{{\left[g(x)\right]}^2}\\ \text{At}\text{x=0:}\\ {\left[\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]\right]}_{x=0}=\frac{g(0)\cdot f\text{'}\left(0\right)-f\left(0\right)\cdot g\text{'}(0)}{{\left[g(0)\right]}^2}\\ =\frac{-3(-2)-(-1)(4)}{{(-3)}^2}\\ =\frac{10}{9}\end{array}$

Conclusion:

The derivative is $\frac{10}{9}$ .

e.

To determine

To calculate: The first derivative of the function at a given point

Answer to Problem 68RE

The derivative is $-2$ .

Explanation of Solution

Given:

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ -1& 0& -1& 2& 1\\ 0& -1& -3& -2& 4\end{array}$

Concept Used:

  $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{\frac{d}{dx}\left[u\left(x\right)\right]\cdot v(x)-u(x)\cdot \frac{d}{dx}\left[v(x)\right]}{v{(x)}^2}$ .

Calculation:

  $\begin{array}{c}\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{\frac{d}{dx}\left[f\left(x\right)\right]\cdot g(x)-f(x)\cdot \frac{d}{dx}\left[g(x)\right]}{{\left[g(x)\right]}^2}\\ =\frac{g(x)\cdot f\text{'}\left(x\right)-f\left(x\right)\cdot g\text{'}(x)}{{\left[g(x)\right]}^2}\\ \text{At}x=-1\text{:}\\ {\left[\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]\right]}_{x=0}=\frac{g(-1)\cdot f\text{'}\left(-1\right)-f\left(-1\right)\cdot g\text{'}(-1)}{{\left[g(-1)\right]}^2}\\ =\frac{-1(2)-(0)(-1)}{{(-1)}^2}\\ =-2\end{array}$

Conclusion:

The derivative is $-2$ .

f.

To determine

To calculate: The first derivative of the function at a given point.

Answer to Problem 68RE

The derivative is $\frac{2}{3}$ .

Explanation of Solution

Given:

  $\begin{array}{ccccc}x& f\left(x\right)& g\left(x\right)& f^{\prime }\left(x\right)& g^{\prime }\left(x\right)\\ -1& 0& -1& 2& 1\\ 0& -1& -3& -2& 4\end{array}$

Concept Used:

  $\frac{d}{dx}\left[u\cdot v\right]=u\frac{d}{dx}\left[v\right]+v\frac{d}{dx}\left[u\right]$ .

Calculation:

  $\begin{array}{c}\frac{d}{dx}\left[\frac{f(x)}{g(x)+2}\right]=\frac{\frac{d}{dx}\left[f\left(x\right)\right]\cdot [g(x)+2]-f(x)\cdot \frac{d}{dx}\left[g(x)+2\right]}{{\left[g(x)\right]}^2}\\ =\frac{[g(x)+2]\cdot f\text{'}\left(x\right)-f\left(x\right)\cdot \left[g\text{'}(x)+0\right]}{{\left[g(x)\right]}^2}\\ \text{At}x=0\text{:}\\ {\left[\frac{d}{dx}\left[\frac{f(x)}{g(x)+2}\right]\right]}_{x=0}=\frac{[g(0)+2]\cdot f\text{'}\left(0\right)-f\left(0\right)\cdot [g\text{'}(0)+0]}{{\left[g(0)\right]}^2}\\ =\frac{\left[-3+2\right]\left(-2\right)-(-1)4}{{(-3)}^2}\\ =\frac{6}{9}\\ =\frac{2}{3}\end{array}$

Conclusion:

The derivative is $\frac{2}{3}$ .