Q. 71 - Calculus

Question

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

$f\left(x\right)=\left\{\begin{array}{l}-x^2, if x\le 0\\ x^2, if x>0\end{array}\right.$

Step-by-Step Solution

Answer

The derivative of the function is $f\text{'}\left(x\right)=\left\{\begin{array}{l}-2x, if x\le 0\\ 2x, if x>0\end{array}\right.$.

Step 1. Given Information

The given function is $f\left(x\right)=\left\{\begin{array}{l}-x^2, if x\le 0\\ x^2, if x>0\end{array}\right.$.

Step 2. Find the derivative

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

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

• Find the derivative for $x>0$.

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

• Since $-2\left(0\right)=2\left(0\right)$, the derivatives at left and right pieces are equal at $x=0$. The derivative exists at $x=0$.
• So, the derivative of the function is, $f\text{'}\left(x\right)=\left\{\begin{array}{l}-2x, if x\le 0\\ 2x, if x>0\end{array}\right.$.