How do I calculate the derivative of $x|x|$? I know that $$f(x)=x\cdot|x|$$ have no derivative at $$x=0$$ but how do I calculate it's derivative for the rest of the points?
When I  calculate for $$x>0$$ I get that $$f'(x) = 2x $$
but for $$ x < 0 $$ I can't seem to find a way to solve the limit. As this is homework please don't put the answer straight forward.
 A: First, draw a picture. Second, make a guess. Third, confirm your guess.

Hmm, looks like $x \mapsto x^2$ for $x \geq 0$ and $x \mapsto -x^2$ for $x \leq 0$. Indeed $f(x) = x^2$ for $x >0 $, so $f'(x) = 2 x$ when $x > 0$. Similarly, $f(x) = -x^2$ when $x <0$, so 
$f'(x) = -2x$ for $x <0$.
The only issue is $x=0$. Looking at the picture, and indeed, looking at the limiting value for $x \downarrow 0$ and $x \uparrow 0$ suggests that if $f$ is differentiable at $x=0$, then $f'(0) = 0$.
So, working with this guess, we try
$$|f(x)-f(0) - 0.x| = |(x|x|)-0 -0.x| = |x|^2$$
If $x\neq 0$, we have $\frac{|f(x)-f(0) - 0.x|}{|x|} = |x|$, from which it follows that 
$f'(0) = 0$. Formally.
To wrap up,  notice that $f'(x) = 2|x|$.
Working without the intuition is difficult. For me, drawing a rough picture provides much of the intuition. If the intuition is sound, then it is often straightforward to formalize the result.
A: There's a great trick that you can use. Notice that for all real $x \neq 0$ you have $|x| \equiv \sqrt{x^2}$.
If you want to know the derivative of $x|x|$ away from $x=0$ then consider $x\sqrt{x^2}$. Using the chain rule and the product rule we can show that, assuming $x \neq 0$,
$$\frac{d}{dx}x|x| \equiv \frac{d}{dx} x\sqrt{x^2} = \sqrt{x^2} + \frac{x^2}{\sqrt{x^2}} \equiv |x| + \frac{x^2}{|x|} \, . $$
Moreover, this little trick works in many more situations. For example, consider $f(|x|)$. Provided we stay away from $x = 0$, $g(|x|) \equiv g(\sqrt{x^2})$ and so:
$$\frac{d}{dx} g(|x|) = \frac{x}{|x|}\frac{d}{dx}g(|x|) \, . $$
A: When $x<0$ replace $|x|$ by $-x$ (since that is what it is equal to) in the formula for the function and proceed. 
Please note as well that the function $f(x)=x\cdot |x|$ does have a derivative at $0$.
A: Hint If $x\gt 0$ we are looking at $x^2$, easy. If $x\lt 0$, we are looking at $-x^2$, easy. For $x=0$, use the definition of the derivative.
So at $0$ we want $\displaystyle\lim_{h\to 0}\,\dots$,
A: *

*For $x \ge 0$ you have $f(x)=x \times |x| = x \times x = x^2$

*For $x \le 0$ you have $f(x)=x \times |x| = x \times (-x) = -x^2$
so you can calculate the derivative when $x \gt 0$ and the derivative when $x \lt 0$ in the usual way.
You are wrong when you say there is no derivative when $x=0$.
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