Find the derivatives of each of the absolute value and piecewise-defined functions in Exercises 65-72.

$f\left(x\right)=\left|x^2-1\right|$

The given function is$f\left(x\right)=\left|x^2-1\right|$.

Rewrite the given function.

$\begin{array}{rcl}f\left(x\right)& =& \left\{\begin{array}{l}-(x^2-1),if{x}^2\le 1\\ x^2-1,if{x}^2>1\end{array}\right.\\ & =& \left\{\begin{array}{l}-(x^2-1),if-1\le x\le 1\\ x^2-1,ifx>1,x<-1\end{array}\right.\\ & =& \left\{\begin{array}{l}{x}^2-1,ifx<-1\\ -(x^2-1),if-1\le x\le 1\\ x^2-1,ifx>1\end{array}\right.\end{array}$

• It is known that, the derivative of the quadratic function is $\left(x^2\right)\text{'}=2x$and the derivative of the constant is 0.
• Find the derivative for $x<-1$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}(x^2-1)\\ & =& 2x\end{array}$

• Find the derivative for $-1\le x\le 1$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}(-(x^2-1\left)\right)\\ & =& -2x\end{array}$

• Since $2(-1)\ne -2(-1)$, the derivatives at left and right pieces are not equal at $x=-1$. The derivative does not exists at $x=-1$.
• Find the derivative for $x>1$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}(x^2-1)\\ & =& 2x\end{array}$

• Since $-2\left(1\right)\ne 2\left(1\right)$, the derivatives at left and right pieces are not equal at $x=1$. The derivative does not exists at $x=1$.
• So, the derivative of the function is, $f\text{'}\left(x\right)=\left\{\begin{array}{l}2x,ifx<-1\\ DNE,ifx=-1\\ -2x,if-1<x<1\\ DNE,ifx=1\\ 2x,ifx>1\end{array}\right.$.