Chapter 2, Problem 67RE

(a).

To determine

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

Answer to Problem 67RE

First derivative of $3f(x)$ at $x=1$ is 12.

Explanation of Solution

Given:

  $\begin{array}{ccc}x& f\left(x\right)& f^{\prime }\left(x\right)\\ 0& 9& -2\\ 1& -3& 4\end{array}$

Concept Used:

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

Calculation:

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

From the given table:

  $\begin{array}{c}\text{At}x=1,\\ 3f\text{'}\left(1\right)=3\times 4\\ =12\end{array}$

b.

To determine

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

Answer to Problem 67RE

Derivative is 1

Explanation of Solution

Given:

  $\begin{array}{ccc}x& f\left(x\right)& f^{\prime }\left(x\right)\\ 0& 9& -2\\ 1& -3& 4\end{array}$

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:

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

Conclusion:

Derivative is 1.

c.

To determine

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

Answer to Problem 67RE

The derivative is $-2$ .

Explanation of Solution

Given:

  $\begin{array}{ccc}x& f\left(x\right)& f^{\prime }\left(x\right)\\ 0& 9& -2\\ 1& -3& 4\end{array}$

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:

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

Conclusion:

Derivative is $-2$ .

d.

To determine

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

Answer to Problem 67RE

The derivative is 7

Explanation of Solution

Given:

  $\begin{array}{ccc}x& f\left(x\right)& f^{\prime }\left(x\right)\\ 0& 9& -2\\ 1& -3& 4\end{array}$

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:

  $\begin{array}{c}\frac{d}{dx}\left[\frac{f(x)}{x}\right]=\frac{\frac{d}{dx}\left[f\left(x\right)\right]\cdot x-f(x)\cdot \frac{d}{dx}\left[x\right]}{{(x)}^2}\\ =\frac{x\cdot f\text{'}\left(x\right)-f\left(x\right)}{{\left(x\right)}^2}\\ \text{At}\text{x=1:}\\ {\left[\frac{d}{dx}\left[\frac{f(x)}{x}\right]\right]}_{x=1}=\frac{1\cdot f\text{'}\left(1\right)-f\left(1\right)}{{\left(1\right)}^2}\\ =\frac{1\times 4+3}{1}\\ =7\end{array}$

Conclusion:

The derivative is 7

e.

To determine

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

Answer to Problem 67RE

The derivative is $-1$ .

Explanation of Solution

Given:

  $\begin{array}{ccc}x& f\left(x\right)& f^{\prime }\left(x\right)\\ 0& 9& -2\\ 1& -3& 4\end{array}$

Calculation:

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

Conclusion:

The derivative is $-1$ .

f.

To determine

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

Answer to Problem 67RE

The derivative is $-36$

Explanation of Solution

Given:

  $\begin{array}{ccc}x& f\left(x\right)& f^{\prime }\left(x\right)\\ 0& 9& -2\\ 1& -3& 4\end{array}$

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:

  $\begin{array}{c}\frac{d}{dx}\left[f(x)\cdot f(x)\right]=\frac{d}{dx}\left[f\left(x\right)\right]\cdot f(x)+f(x)\cdot \frac{d}{dx}\left[f(x)\right]\\ =f\text{'}(x)\cdot f(x)+f(x)\cdot f\text{'}(x)\\ \text{At}\text{x=0:}\\ {\left[\frac{d}{dx}\left[f\left(x\right)\cdot f(x)\right]\right]}_{x=0}=f\text{'}(0)\cdot f(0)+f(0)\cdot f\text{'}(0)\\ =2\times \left(-2\right)\times 9\\ =-36\end{array}$

Conclusion:

The derivative is $-36$