$$\mathop {\lim }\limits_{x \to {{{\pi ^ - }} \over 2}} f(x) = 0 = f( - \pi /2)$$

$$\mathop {\lim }\limits_{x \to {{{\pi ^ + }} \over 2}} f(x) = \cos \left( { - {\pi \over 2}} \right) = 0$$

$$f'(x) = \left\{ {\matrix{ { - 1,} & {x \le - \pi /2} \cr {\sin x,} & { - \pi /2 < x \le 0} \cr {1,} & {0 < x \le 1} \cr {1/x,} & {x > 1} \cr } } \right.$$

Clearly, f(x) is not differentiable at x = 0 as f'(0^$$-$$) = 0 and f'(0⁺) = 1.

f(x) is differentiable at x = 1 as $$f'({1^{ - 1}}) = f'({1^ + }) = 1$$.