Check Differentiability chech whether the function is differentiable at $x=0$
$$f(x)=\left\lbrace \begin{array}{cl}
\arctan\frac{1}{\left | x \right |}, & x\neq 0 \\ 
\frac{\pi}{2}, &  x=0\\ 
 \end{array}\right.$$
I feel that this is differentable at given point but I am unable to proceed with it.
 A: As suggested in comments, if $x \not =0$, $\frac{f(x)-f(0)}{x-0}=\frac{-\arctan|x|}{x}$. Now if $x \to 0^{+}$, then $\lim\limits_{x \to 0^{+}}\frac{-\arctan|x|}{x}=\lim\limits_{x \to 0^{+}}\frac{-\arctan x}{x}=\lim\limits_{x \to 0^{+}}\frac{-\arctan x - (-\arctan(0)}{x-0}=(-\arctan)'(0)=-\frac{1}{0^{2}+1}=-1$, Now as $x \to 0^{-}$, $\lim\limits_{x \to 0^{-}}\frac{-\arctan|x|}{x}=\lim\limits_{x \to 0^{-}}\frac{-\arctan(-x)}{x}=1$ (Same way as before, and we use the  chain rule, the derivative of $k(x)=\arctan(-x)$ is $\frac{-1}{x^{2}+1}$.) The limit is different at both sides, so $f$ is not differentiable at $0$
A: Simply use the identity
$$\arctan (1/x)=\frac{\pi}{2}\text{sgn}(x)-\arctan(x)$$
Then, the limit of the difference quotient is 
$$\lim_{h\to0}\frac{\arctan(1/|h|)-\frac{\pi}{2}}{h}=-\lim_{h\to0}\frac{\arctan(|h|)}{h} \tag 1$$
We see that the limit in $(1)$ does not exist since the limit from the right side does not equal the limit from the left side.  In fact, a quick check using say L'Hospital's Rule shows that the limit from the right side is $-1$, while the limit from the left side is $1$.
We conclude, therefore, that $\arctan(1/|x|)$ is not differentiable at $x=0$.
