What is the derivative of \( f(x)=x|x| \) at \( x=0 \) ? Choose the correct answer below. A. The derivative does not exist because \( |x| \) is not differentiable at \( x=0 \). B. The derivative does not exist because the left- and right-hand limits do not agree. C. The derivative does not exist because \( f \) is defined piecewise. D. The derivative is

We start by rewriting the function \( f(x) = x|x| \) as a piecewise function. Note that \[ |x|=\begin{cases} x & \text{if } x\ge0, \\ -x & \text{if } x<0. \end{cases} \] Thus, \[ f(x)=\begin{cases} x\cdot x=x^2 & \text{if } x\ge0, \\ x\cdot(-x)=-x^2 & \text{if } x<0. \end{cases} \] Next, we compute the derivative at \( x=0 \) using the definition of the derivative: \[ f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}. \] Since \( f(0)=0 \) (because \( 0|0|=0 \)), the expression simplifies to: \[ f'(0)=\lim_{h\to0}\frac{f(h)}{h}. \] We now consider the limit from the right (\( h>0 \)) and from the left (\( h<0 \)). 1. **For \( h>0 \):** When \( h>0 \), \( f(h)=h^2 \). Thus, \[ \frac{f(h)}{h}=\frac{h^2}{h}=h. \] Taking the limit as \( h\to0^+ \), \[ \lim_{h\to0^+}h=0. \] 2. **For \( h<0 \):** When \( h<0 \), \( f(h)=-h^2 \). Thus, \[ \frac{f(h)}{h}=\frac{-h^2}{h}=-h. \] Taking the limit as \( h\to0^- \), \[ \lim_{h\to0^-}(-h)=0. \] Since both one-sided limits exist and equal \( 0 \), the derivative at \( x=0 \) is \[ f'(0)=0. \] Thus, among the given options, the correct answer is: D. The derivative is \( 0 \).