Chapter 4.1, Problem 78E

(a)

To determine

To calculate: To show that $\frac{d}{dx}\left|u\right|=u\text{'}\frac{u}{\left|u\right|}$ .

Answer to Problem 78E

Hence showed that $\frac{d}{dx}\left|u\right|=u\text{'}\frac{u}{\left|u\right|}$

Explanation of Solution

Given Information: u is a differentiable function of x.

Calculation:

The derivative of the function is,

  $\begin{array}{c}\left|u\right|=\sqrt{u^2}\\ \frac{d}{dx}\left|u\right|=\frac{d}{dx}\sqrt{u^2}\\ =\frac{1}{2\sqrt{u^2}}\frac{d}{dx}\left(u^2\right)\left(\text{Apply chain rule; Use }\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\right)\\ =\frac{1}{2\sqrt{u^2}}\times 2u\frac{du}{dx}\left(\text{Write the values of the derivative}\right)\\ =\frac{u}{\sqrt{u^2}}\frac{du}{dx}\left(\text{Simplify}\right)\\ =\frac{u}{\sqrt{u^2}}u\text{'}\left(\frac{du}{dx}=u\text{'}\right)\\ =\frac{u}{\left|u\right|}u\text{'}\left(\sqrt{u^2}=\left|u\right|\right)\end{array}$

(b)

To determine

To calculate: Find the derivative of $f\left(x\right)=\left|x^2-9\right|$ and $g\left(x\right)=\left|x\right|\sin x$

Answer to Problem 78E

The derivative of the function $f\left(x\right)=\left|x^2-9\right|$ is $\frac{2x\left(x^2-9\right)}{\left|x^2-9\right|}$ .

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

Explanation of Solution

Given Information: $\frac{d}{dx}\left|u\right|=u\text{'}\frac{u}{\left|u\right|}$

Calculation:

The derivative of the function $f\left(x\right)=\left|x^2-9\right|$ is,

  $\begin{array}{c}f\left(x\right)=\left|x^2-9\right|\\ \frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left|x^2-9\right|\left(\text{Substitute the value }f\left(x\right)=\left|x^2-9\right|\right)\\ =\frac{d}{dx}\left|u\right|\left(\text{let }{x}^2-9=u\right)\\ =\frac{u}{\left|u\right|}u\text{'}\left(\because \frac{d}{dx}\left|u\right|=\frac{u}{\left|u\right|}u\text{'}\right)\\ \text{=}\frac{x^2-9}{\left|x^2-9\right|}\frac{d}{dx}\left(x^2-9\right)\left(\text{Substitute the value of }u\right)\\ =\frac{x^2-9}{\left|x^2-9\right|}\times 2x\\ =\frac{2x\left(x^2-9\right)}{\left|x^2-9\right|}\end{array}$

The derivative of the function $f\left(x\right)=\left|x^2-9\right|$ is $\frac{2x\left(x^2-9\right)}{\left|x^2-9\right|}$ .

The derivative of the function $g\left(x\right)=\left|x\right|\sin x$ is,

  $\begin{array}{c}g\left(x\right)=\left|x\right|\sin x\\ \frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(\left|x\right|\sin x\right)\left(\text{Substitute the value }f\left(x\right)=\left|x^2-9\right|\right)\\ =\sin x\frac{d}{dx}\left|x\right|\text{+}\left|x\right|\frac{d}{dx}\left(\text{sinx}\right)\left(\text{Apply chain rule}\right)\\ =\sin x\frac{d}{dx}\left|u\right|\text{+}\left|x\right|\cos x\left(\text{let }x=u\right)\\ =\sin x\frac{u}{\left|u\right|}u\text{'}\text{+}\left|x\right|\cos x\left(\because \frac{d}{dx}\left|u\right|=\frac{u}{\left|u\right|}u\text{'}\right)\\ \text{=}\sin x\frac{x}{\left|x\right|}\frac{d}{dx}\left(x\right)\text{+}\left|x\right|\cos x\left(\text{Substitute the value of }u\right)\\ \text{=}\sin x\frac{x}{\left|x\right|}\text{+}\left|x\right|\cos x\\ =\frac{x\sin x}{\left|x\right|}\text{+}\left|x\right|\cos x\end{array}$

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