Does the derivative of $f(x) = 2x^2 |x|$ exist at $x=0$? Maybe an elementary question. I'm helping some kids I know study for their calculus exam, and somehow I'm confused about this and I can't figure out a way to explain the answer to them? So, from what I think, if you take the derivative for values greater than zero and less than zero, you get:
$$\frac{df}{dx} = 6x^2$$ and
$$\frac{df}{dx} = -6x^2$$
respectively, right?
But, is this differentiable at $x=0$? Wouldn't the limits of both the sides be different?
 A: That function is differentiable at $0$; the derivative there is $0$. You can see that clearly in a picture.
To prove it from the definition of the derivative, note that for
$h \ne 0$ the difference quotient is
$$
\frac{f(0+h) -f(0)}{h} = \frac{2h^2|h|}{h} = 2h|h|
$$
which has limit $0$ as $h$ approaches $0$.
In fact, the derivative as a function is itself differentiable. That's not true for $f(x) = x|x|$.
A: Yes.
Let $f(x)=2x^2\left| x \right|$.
Since $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}2x\left| x\right| = 0$$
exist, $f(x)$ is differentiable at $x=0$.
A: You seem to misunderstand how differentiability works. We have two functions $f,$ $g,$ defined by $f(x)=|x|$ and $g(x)=2x^2|x|$ respectively, so that $g(x)=2x^2f(x).$ You are correct in pointing out that $f$ is not differentiable at $0.$ However, that does not imply $g$ is not differentiable either. I do not know why you are making this leap in logic. The only requirement for $g$ to be differentiable at $0$ is for $$\lim_{x\to0}\frac{g(x)-g(0)}{x}$$ to exist. Whether $f$ is differentiable at $0$ or not is not relevant. We know that $g(0)=0,$ so $$g'(0)=\lim_{x\to0}\frac{g(x)}{x}=\lim_{x\to0}2xf(x).$$ So as long as this latter limit exists, the differentiability, or lack thereof, of $f$ at $0$ does not matter. Does this limit exist? That is the only question you need to answer to determine whether $g$ is differentiable at $0$ or not.
I suspect the reason you are getting so caught up on the non-differentiability of $f$ at $0$ is because it seems to me you are thinking that the product $f\cdot{g}$ can only be differentiable at $0$ if both $f$ and $g$ are differentiable at $0.$ That is just not true, though. Suppose $f(0)=0$ and $g(0)=0.$ Then $$(f\cdot{g})'(x)=\lim_{x\to0}\frac{f(x)g(x)}{x}$$ must exist. Now, if $f$ and $g$ are differentiable, then it certainly is true that the limit exists. But such a strong condition is not necessary for the limit to exist. It only requires that one of the two functions be differentiable at $0,$ and that the limit for the other function exist. If $g$ is differentiable at $0,$ then $$\lim_{x\to0}\frac{g(x)}{x}$$ must exist. So if $$\lim_{x\to0}f(x)$$ also exists, then that is more than sufficient for $$\lim_{x\to0}\frac{f(x)g(x)}{x}=\left[\lim_{x\to0}f(x)\right]\left[\lim_{x\to0}\frac{g(x)}{x}\right]=g'(0)\lim_{x\to0}f(x)$$ to exist and be true.
