Is $f(x)=|x-1|^3$ differentiable $\forall x$ Is $f(x)=|x-1|^3$ differentiable $\forall x$?
I can't understand if even if there is no cusp there is still a problem in differentiating it in $x=1$. 
 A: Unpack the definition of the absolute value a bit:
$$
\lvert x \rvert = \begin{cases}
\hfil x, & x \geq 0, \\
-x, & x < 0.
\end{cases}
$$
Then your function can be presented in a similar piecewise form:
$$
f(x) = \lvert x-1 \rvert^3 = \begin{cases}
\hfil (x-1)^3, & x \geq 1, \\
-(x-1)^3, & x < 1.
\end{cases}
$$
This function is clearly differentiable for all $x \neq 1$.  But, at $x=1$, there's a potential problem.  You want to consider the derivative on either side and look at the limits as $x \to 1$.  Notice that we deliberately exclude writing down $f'(1)$ for the moment, since it might not exist.
$$
f'(x) = \begin{cases}
\hfil 3(x-1)^2, & x > 1, \\
-3(x-1)^2, & x < 1.
\end{cases}
$$
So,
$$
\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} 3(x-1)^2 = 0
$$
and
$$
\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} -3(x-1)^2 = 0
$$
Since, they're equal, we can define
$$
f'(1) = 0
$$
and $f$ is differentiable for all real $x$.
One way to write this back in a single formula is
$$
f'(x) = 3 \lvert x-1 \rvert (x-1)
$$

By the way, a similar argument shows that $f$ is twice differentiable (i.e., $f''(x)$ exists for all $x$; in fact you can write $f''(x) = 6\lvert x-1 \rvert$), but that's it.  You can check by looking at limits from the right and left that $f'''(1)$ does not exist.
A: Indeed the function is differentiable at all $x\neq 1$. The possible problem is with $1$.
So, try the definition of the derivative:
$$
f'(x) = \lim_{x \to 1} \frac{f(x) - f(1)}{x-1} = \lim_{x \to 1} \frac{\lvert x-1\rvert^3}{x-1}
$$
The function is differentiable at $1$ exactly if this limit exists. So try to check if the left-hand limit and the right-hand limit are the same. For example
$$
\lim_{x \to 1^+} \frac{\lvert x-1\rvert^3}{x-1} = \lim_{x \to 1^+} \frac{(x-1)^3}{x-1} = \dots
$$
And 
$$
\lim_{x \to 1^-} \frac{\lvert x-1\rvert^3}{x-1} = \lim_{x \to 1^-} \frac{-(x-1)^3}{x-1} = \dots
$$
A: No, there is not:
$$ \lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{|h|^3}{h}=\lim_{h\to0}h^2\operatorname{sgn}(h)=0.$$
It turns out that $f'(x)=3\left|x-1\right|(x-1)$ and $f''(x)=6\left|x-1\right|$ so that $f$ is twice continuously differentiable!
