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$$ f(x) = \begin{cases} \arctan {x}, & \text{if $|x| \leq 1$} \\[2ex] \dfrac{\pi x}{4 |x|} + \dfrac{|x|-1}{2}& \text{if $|x| > 1$} \end{cases} $$ I think the function is continuous only where the values coincide. But, don't know about differentiability.

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Continuity is a problem only at $1$ and $-1$. However, $$ \lim_{x\to1^+}f(x)= \lim_{x\to 1^+}\left(\frac{\pi x}{4x}+\frac{x-1}{2}\right)=\frac{\pi}{4} $$ and the same for the limit $x\to 1^-$. Similarly at $-1$.

For $01$, $$ f'(x)=\frac{1}{2} $$ Therefore $$ \lim_{x\to1}f'(x)=\frac{1}{2} $$ Apply l‘Hôpital.

What happens at $-1$?

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    I can't understand why continuity is a problem at 1 and -1 – 2017-01-14
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    @Infinity It's where the definition “changes”. Any other point has a neighborhood where the function is (the restriction of) a continuous function. – 2017-01-14