Chapter 3.4, Problem 92E

a.

To determine

Explain the proof of the given relation using chain rule.

Explanation of Solution

Given:

The given relationis $\frac{d}{dx}\left|x\right|=\frac{x}{\left|x\right|}$ , where $\left|x\right|=\sqrt{x^2}$ .

Calculation:

  $\frac{d}{dx}\left|x\right|=\frac{x}{\left|x\right|}$

From L.H.S

  $\frac{d}{dx}\left|x\right|$

  $\left|x\right|=\sqrt{x^2}$

  $\frac{d}{dx}\left|x\right|=\frac{d}{dx}\sqrt{x^2}$

Apply chain rule.

  $\frac{df\left(a\right)}{dx}=\frac{df}{da}\cdot \frac{da}{dx}$

Let $f=\sqrt{a},a=x^2$

  $\Rightarrow \frac{d}{dx}\left|x\right|=\frac{d}{da}\left(\sqrt{a}\right)\cdot \frac{d}{dx}\left(x^2\right)$

Use derivative rule.

  $\frac{d}{dx}\left(\sqrt{x}\right)=\frac{1}{2\sqrt{x}}\text{and}\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{l}\Rightarrow \frac{d}{dx}\left|x\right|=\frac{1}{2\sqrt{a}}\cdot 2x\\ \Rightarrow \frac{d}{dx}\left|x\right|=\frac{x}{\sqrt{a}}\end{array}$

Substitute the value of $\left|x\right|=\sqrt{x^2}$ .

  $\begin{array}{l}\Rightarrow \frac{d}{dx}\left|x\right|=\frac{x}{\sqrt{x^2}}\\ \Rightarrow \frac{d}{dx}\left|x\right|=\frac{x}{\left|x\right|}=\text{R}\text{.H}\text{.S}\end{array}$

Hence $\frac{d}{dx}\left|x\right|=\frac{x}{\left|x\right|}$ proved.

b.

To determine

Find the derivative of the given function and sketch the graph of function and its derivative.

Answer to Problem 92E

The derivative of the function is $f\text{'}\left(x\right)=\frac{\sin x\cos x}{\left|\sin x\right|}$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=\left|\sin x\right|$ , where $\left|x\right|=\sqrt{x^2}$ .

Calculation:

  $f\left(x\right)=\left|\sin x\right|$

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

where $\left|\sin x\right|=\sqrt{{\sin }^{2}x}$ .

Apply chain rule.

  $\frac{df\left(a\right)}{dx}=\frac{df}{da}\cdot \frac{da}{dx}$

Let $f=\sqrt{a},a={\sin }^{2}x$

  $\Rightarrow f\text{'}\left(x\right)=\frac{d}{da}\left(\sqrt{a}\right)\cdot \frac{d}{dx}\left({\sin }^{2}x\right)$

Use derivative rule.

  $\frac{d}{dx}\left(\sqrt{x}\right)=\frac{1}{2\sqrt{x}},\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\text{and}\frac{d}{dx}\left(\sin x\right)=\cos x$ .

  $\begin{array}{l}\Rightarrow f\text{'}\left(x\right)=\frac{1}{2\sqrt{a}}\cdot 2\sin x\cos x\\ \Rightarrow f\text{'}\left(x\right)=\frac{\sin x\cos x}{\sqrt{a}}\end{array}$

Substitute the value of $\left|\sin x\right|=\sqrt{{\sin }^{2}x}$ .

  $\begin{array}{l}\Rightarrow f\text{'}\left(x\right)=\frac{\sin x\cos x}{\sqrt{{\sin }^{2}x}}\\ \Rightarrow f\text{'}\left(x\right)=\frac{\sin x\cos x}{\left|\sin x\right|}\end{array}$

The graph of the function and its derivative is given below.

  

Green line shows the graph of the function and blue line shows the graph of its derivative.

Hence the derivative of the function is $f\text{'}\left(x\right)=\frac{\sin x\cos x}{\left|\sin x\right|}$ .

c.

To determine

Find the derivative of the given function and sketch the graph of function and its derivative.

Answer to Problem 92E

The derivative of the function is $g\text{'}\left(x\right)=\frac{x\cos \left|x\right|}{\left|x\right|}$ .

Explanation of Solution

Given:

The given function is $g\left(x\right)=\sin \left|x\right|$ , where $\left|x\right|=\sqrt{x^2}$ .

Calculation:

  $g\left(x\right)=\sin \left|x\right|$

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

where $\left|x\right|=\sqrt{x^2}$ .

Apply chain rule.

  $\frac{df\left(a\right)}{dx}=\frac{df}{da}\cdot \frac{da}{dx}$

Let $f=\sin a,a=\sqrt{x^2}$

  $\Rightarrow g\text{'}\left(x\right)=\frac{d}{da}\left(\sin a\right)\cdot \frac{d}{dx}\left(\sqrt{x^2}\right)$

Use derivative rule.

  $\frac{d}{dx}\left(\sqrt{x}\right)=\frac{1}{2\sqrt{x}},\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\text{and}\frac{d}{dx}\left(\sin x\right)=\cos x$ .

  $\begin{array}{l}\Rightarrow g\text{'}\left(x\right)=\cos a\cdot \frac{2x}{2\sqrt{x^2}}\\ \Rightarrow g\text{'}\left(x\right)=\cos a\cdot \frac{x}{\sqrt{x^2}}\end{array}$

Substitute the value of $\left|x\right|=\sqrt{x^2}$ .

  $\begin{array}{l}\Rightarrow g\text{'}\left(x\right)=\cos \sqrt{x^2}\cdot \frac{x}{\sqrt{x^2}}\\ \Rightarrow g\text{'}\left(x\right)=\frac{x\cos \left|x\right|}{\left|x\right|}\end{array}$

The graph of the function and its derivative is given below.

  

Green line shows the graph of the function and blue line shows the graph of its derivative.

Hence the derivative of the function is $g\text{'}\left(x\right)=\frac{x\cos \left|x\right|}{\left|x\right|}$ .