How is $f(x) = x^2|x|$ differentiable for $c > 0$?

Question

What I've gotten so far is the following.

Let $f(x) = x^2|x|$. Prove that $f$ is differentiable at every point. $\lim_{x \to 0} \frac{f(x)-f(0)}{x-0} =\lim_{x \to 0} \frac{x^{2}|x|}{x} = \lim_{x \to 0}x|x|$ = $x^2 \rightarrow 0$ as $x \rightarrow 0$. So, $f(x)$ is differentiable at $0$.

I was wondering how I prove that $f(x)$ is differentiable for $c > 0$ ?

For $x > 0$, $f(x) = x^2 \cdot x = x^3$. For $x<0$, $f(x) = -x^3$. — Math Lover, Nov 19 '17 at 15:45
The fact that $x^3$ is differentiable follows pretty quickly from the limit definition. — Stella Biderman, Nov 19 '17 at 15:47
@Thomas, in my notes it says $c > 0$ even though $x$ is used. — Skm, Nov 19 '17 at 15:48
I wanted to prove it showing that $\lim{x \to c}\frac{x^3-c^3}{x-c}$ and then simplify from there, but I'm having a hard time doing that. — Skm, Nov 19 '17 at 15:56
@Sharon: https://math.stackexchange.com/questions/730596/simplifying-the-expression-fracx3-y3x-y, https://math.stackexchange.com/questions/1789286/how-to-factor-a3-b3. — Martin R, Nov 19 '17 at 16:02
There is a mistake in your proof. $x|x| = \pm x^2$ depending on whether $x>0$ or $x<0$. — Ennar, Nov 20 '17 at 05:46

score 2 · Answer 1 · answered Nov 19 '17 at 15:47

2

Because for all $x>0$, we have $$(x^2|x|)'=(x^3)'=3x^2.$$

answered Nov 19 '17 at 15:47

Michael Rozenberg

203,855

score 1 · Accepted Answer · answered Nov 19 '17 at 16:09

1

As I understand, you are already aware of the fact that for $x>0$, $f(x)=x^3$, but you have problems showing that this is differentiable.

We have $$f'(c) = \lim_{x \to c} \frac {f(x)-f(c)} {x-c} = \lim_{h \to 0} \frac {f(c+h)- f(c)} h\\ =\lim_{h \to 0} \frac{(c+h)^3-c^3}{h} = \lim_{h\to 0} \frac {3hc^2 +3h^2c+h^3}{h} \\ \lim_{h\to 0}3c^2 +3hc +h^2 = 3c^2 $$

Thus $f$ is differentiable for $x>0$ and has the derivative $f'(x)=3x^2$.

Can you show that $f$ is differetiable for $x<0$?

answered Nov 19 '17 at 16:09

klirk

3,724

Yes, this helped a lot. I now understand how to show that $f$ is differentiable for $x < 0$. – Skm Nov 19 '17 at 16:11
So what is the derivative? – klirk Nov 19 '17 at 16:12
for $x<0$ the derivative is $-3x^2$ – Skm Nov 19 '17 at 16:13

How is $f(x) = x^2|x|$ differentiable for $c > 0$?

2 Answers2