Differentiability of $x|x|$

Detailed Solution

For $x>0$, $$ |x|=x, $$ so $$ f(x)=x^2. $$ Hence, $$ f'(x)=2x,\qquad x>0. $$ For $x<0$, $$ |x|=-x, $$ so $$ f(x)=-x^2. $$ Hence, $$ f'(x)=-2x,\qquad x<0. $$ Thus, $f$ is differentiable at every $x\neq0$. It remains to check differentiability at $x=0$. Since $$ f(0)=0, $$ the derivative at $0$ is $$ f'(0) = \lim_{h\to0}\frac{f(0+h)-f(0)}{h}. $$ Using $f(h)=h|h|$, $$ f'(0) = \lim_{h\to0}\frac{h|h|}{h} = \lim_{h\to0}|h| = 0. $$ Therefore, $$ \boxed{f'(0)=0}. $$ We can also verify the one-sided derivatives: $$ f'_-(0) = \lim_{h\to0^-}|h| =0, $$ and $$ f'_+(0) = \lim_{h\to0^+}|h| =0. $$ Hence, $$ f'_-(0)=f'_+(0)=0, $$ so $f$ is differentiable at $0$. In fact, the derivative can be written as $$ f'(x)= \begin{cases} -2x,&x<0,\\ 0,&x=0,\\ 2x,&x>0. \end{cases} $$ Option A is false because $f$ is also differentiable at $0$. Option C is false because the left and right derivatives at $0$ are both $0$. Option D is false because although $|x|$ itself is not differentiable at $0$, the product $x|x|$ can still be differentiable there. Direct evaluation of the derivative gives $f'(0)=0$. Therefore, the correct option is $$ \boxed{\mathrm{B}}. $$