Finding the derivative value at x = 0 of $x|x|$

Question

please help me with this problem,

I want to find the value of the derivative of the function $f(x) = x|x|$ at $x=0$

if I'm using the following limit definition

$$\lim_{\Delta x\to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$$

at $x = 0$ , I get: $$\lim_{\Delta x\to 0} \frac{f(0+\Delta x)-f(0)}{\Delta x} = \lim_{\Delta x\to 0} \frac{\Delta x |\Delta x|}{\Delta x} =\lim_{\Delta x\to 0} |\Delta x| = 0$$ but the derivative of $x|x|$ is $\frac{2x^2}{|x|}$ and for $x=0$ I get $\frac{0}{0}$ which is, of course, undefined.

so what is the derivative of $x|x|$ at $x=0$

Thank you

Answer 1

The function is actually differentiable at $x = 0$ and the derivative is indeed equal to $0$. Note that, $$x|x| = \begin{cases} x^2 & \text{for } x \ge 0 \\ -x^2 & \text{for } x < 0 \end{cases} $$ Thus, $$(x|x|)' = \begin{cases} 2x & \text{for } x \ge 0 \\ -2x & \text{for } x < 0 \end{cases} $$ This clearly gives us, $(x|x|)' = 0$ at $x = 0$. The formula you stated, as said in the comments, only works when $x \ne 0$.