# Thread: Proving x|x| is differentiable

1. ## Proving x|x| is differentiable

I just have a quick, and most likely silly question. I need to show that the function x|--> x|x| is differentiable on R. Would the best way to do this be to let g(x) = x, h(x) = |x|, and use the product rule to show that the derivative is 2x for all x?

Is there anything special I would need to do for x = 0, since |x| isn't differentiable at x=0?

2. Originally Posted by monsieur fatso
I just have a quick, and most likely silly question. I need to show that the function x|--> x|x| is differentiable on R. Would the best way to do this be to let g(x) = x, h(x) = |x|, and use the product rule to show that the derivative is 2x for all x?

Is there anything special I would need to do for x = 0, since |x| isn't differentiable at x=0?

$f(x) = x|x|$

for the case $x = 0$ ...

$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$

$f'(0) = \lim_{x \to 0} \frac{x|x|}{x} = 0$

should be clear that the limit also exists for all $x \neq 0$

3. Okay, I guess I'm having some trouble evaluating that limit..
$f'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x} = \frac{x|x|-x|x|}{x-x}=|x|
$

Which isn't correct...

4. $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

$f'(a) = \lim_{x \to a} \frac{x|x| - a|a|}{x - a}$

for $x$ and $a$ both positive ...

$f'(a) = \lim_{x \to a} \frac{x^2 - a^2}{x - a}
$

$f'(a) = \lim_{x \to a} (x + a) = 2a$

for $x$ and $a$ both negative ...

$f'(a) = \lim_{x \to a} \frac{x(-x) - a(-a)}{x - a}$

$f'(a) = -\lim_{x \to a} \frac{x^2 - a^2}{x - a}$

$f'(a) = -\lim_{x \ to a} (x + a) = -2a
$