Q. 70 - Calculus

Question

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

$f\left(x\right)=\left\{\begin{array}{l}1, if x\le -1\\ x^{2/3}, if x>-1\end{array}\right.$

Step-by-Step Solution

Answer

The derivative of the function is $f\text{'}\left(x\right)=\left\{\begin{array}{l}0, if x<-1\\ DNE, if x=-1\\ \begin{array}{rcl}\frac{2}{3}{x}^{\frac{-1}{3}}, if x& >& -1\end{array}\end{array}\right.$.

Step 1. Given Information

The given function is $f\left(x\right)=\left\{\begin{array}{l}1, if x\le -1\\ x^{2/3}, if x>-1\end{array}\right.$

Step 2. Find the derivative

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(1\right)\\ & =& 0\end{array}$

• Find the derivative for $x>-1$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(x^{2/3}\right)\\ & =& \frac{2}{3}{x}^{\frac{2}{3}-1}\\ & =& \frac{2}{3}{x}^{\frac{-1}{3}}\end{array}$

• Since $0\ne \begin{array}{rcl}& & \frac{2}{3}\end{array}\begin{array}{rcl}& & (\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 1\end{array}\begin{array}{rcl}& & )\end{array}\begin{array}{rcl}& & \frac{-1}{3}\end{array}$, 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, $f\text{'}\left(x\right)=\left\{\begin{array}{l}0, if x<-1\\ DNE, if x=-1\\ \begin{array}{rcl}\frac{2}{3}{x}^{\frac{-1}{3}}, if x& >& -1\end{array}\end{array}\right.$.