Chapter 2, Problem 10RWDT

To determine

To find: To show the derivative of $f(x)=\left\{\begin{array}{l}{x}^3\text{if}x\ge \text{0}\\ -x^3\text{if}x<0\end{array}\right.$ is $f\text{'}\left(x\right)=3x\left|x\right|$ .

Explanation of Solution

Given information:

A function, $f(x)=\left\{\begin{array}{l}{x}^3\text{if}x\ge \text{0}\\ -x^3\text{if}x<0\end{array}\right.$

Concept Used

a. Power rule differentiation: $\frac{d{x}^n}{dx}=n{x}^{n-1}$

b. $\left|x\right|=\left\{\begin{array}{l}-x\text{if}x<0\\ x\text{if}x\ge \text{0}\end{array}\right.$

where $a$ and $b$ are constants $u\left(x\right)$ and $v\left(x\right)$ are functions. of $x$

Proof:

Differentiate $f(x)=\left\{\begin{array}{l}{x}^3\text{if}x\ge \text{0}\\ -x^3\text{if}x<0\end{array}\right.$as follows:

For $x\ge 0$ :

  $\begin{array}{c}\frac{d}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[x^3\right]\\ =3x^2\end{array}$

For $x<0$ :

  $\begin{array}{c}\frac{d}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[-x^3\right]\\ =-3x^2\end{array}$

This can be written as:

  $f\text{'}(x)=\left\{\begin{array}{l}3x^2\text{if}x\ge \text{o}\\ {\text{-3x}}^2\text{if }x\text{<0}\end{array}\right.$

By putting mod and writing it together as follows:

  $f\text{'}(x)=3x\left|x\right|$ .