Is $|9x-1|^3$ differentiable at $1/9$? I think it is, but I'm stuck at showing it. I know that it is equal to:
$$|9x - 1|  \cdot |9x - 1|^2.$$ 
$|9x - 1|$ is not differentiable at $1/9$, and I am pretty sure that $|9x - 1| ^2$ is differentiable there.
 A: Let me give you a more general perspective. Consider a smooth (say $C^1$) function $h$ and define $f(x)=h(|x|)$. By assumption,
$$
h(u) = h(0)+\left( h'(0) + \omega(u) \right)u,
$$
where $\lim_{u \to 0} \omega (u)=0$. Hence
$$
f(x) = h(|0|) + \left( h'(0) + \omega(|x|) \right) |x|
$$
and therefore $f$ is differentiable at $x=0$ if and only if $h'(0)=0$. This is the case for $h(u)=u^3$.
A: Its quite easy and same as checking for $f(x)=|x-1/9|^3$.
For $x>1/9$
$f(x)=(x-1/9)^3$
Differentiate w.r.t x and put $x=1/9$.Say the value of the derivative is $a$.
For $x<1/9$
$f(x)=-(x-1/9)^3$
Differentiate w.r.t x and put $x=1/9$.Say the value of the derivative is $b$.
If $a=b$ then it is differentiable at $x=1/9$
This method works because $f(x)$ is continuous at $x=1/9$
A: We can look separately at the limits of $\frac{f(x)-f(1/9)}{x-1/9}$ (for $x\neq 1/9$) on both sides of $1/9$: the goal is to see if they exist, and (if so) if they are the same. If this is the case, then $f$ is differentiable at $1/9$, and $f^\prime(1/9)$ equals this common limit.
See below for more details. The hidden parts can be revealed by placing your mouse over them.

Start by factorizing $(9x-1)^3$ to make a factor $x-\frac{1}{9}$ appear:

$$(9x-1)^3 = 9\left(x-\frac{1}{9}\right)(9x-1)^2$$



*

*For $x > \frac{1}{9}$, $\lvert 9x-1\rvert^3 = (9x-1)^3$, so that

 $$\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}} = \frac{(9x-1)^3-0}{x-\frac{1}{9}} = 9(9x-1)^2 \xrightarrow[x\to1/9^+]{} 0$$


*For $x < \frac{1}{9}$, $\lvert 9x-1\rvert^3 = (1-9x)^3 = -(9x-1)^3$, so that

 $$\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}} = -\frac{(9x-1)^3-0}{x-\frac{1}{9}} = -9(9x-1)^2 \xrightarrow[x\to1/9^-]{} 0$$

Therefore, the limits on both sides exist and coincide: $f^\prime(1/9)=\lim_{x\to1/9}\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}}$ thus exists.
