Q. 69 - Calculus

Question

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

$f\left(x\right)=\left\{\begin{array}{l}{x}^3, if x<1\\ x, if x\ge 1\end{array}\right.$

Step-by-Step Solution

Answer

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

Step 1. Given Information

The given function is $f\left(x\right)=\left\{\begin{array}{l}{x}^3, if x<1\\ x, if x\ge 1\end{array}\right.$.

Step 2. Find the derivative

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

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

• Find the derivative for $x>1$.

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

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