$f(x) = ||x| - 1|$ is not differentiable at

Let $f:[-1,1] \rightarrow R$ be a function defined by $f(x)=\begin{cases} x^2 \left| \cos \left(\frac{\pi}{x}\right) \right| & \text{for } x \neq 0 \\ 0 & \text{for } x=0 \end{cases}$. The set of points where $f$ is not differentiable is

Let $f : R \to R$ be differentiable at $c \in R$ and $f(c) = 0$. If $g(x) = |f(x)|$,then at $x = c$,$g$ is

If $y = \sec^{-1} \left( \frac{2x}{1 + x^2} \right) + \sin^{-1} \left( \frac{x - 1}{x + 1} \right)$,then $\frac{dy}{dx}$ is equal to

The set of points where the function $f(x)=|x-1| e^{x}$ is differentiable,is

Let $f(x) = \begin{cases} g(x) \cos(\frac{1}{x}) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$ where $g(x)$ is an even function differentiable at $x = 0$,passing through the origin. Then $f'(0)$: