Let f(x) = sin x, g(x) = [x + 1] and g{f(x)} = h(x), where [.] is the greatest integer function. Then $h'(\frac{π}{2})$ is

non existent

$h(x) = g\{f(x)\}=[f(x)+1] =[\sin x +1]$

$h'\left(\frac{π}{2}+0\right)=\underset{h→0}{\lim}\frac{\left[\sin\left(\frac{π}{2}+h\right)+1\right]-\left[\sin\frac{π}{2}+1\right]}{h}=\underset{h→0}{\lim}\frac{[\cos h+1]-[2]}{h}$

$=\underset{h→0}{\lim}\frac{1-2}{h}=\underset{h→0}{\lim}\frac{-1}{h}=-∞$

$h'\left(\frac{π}{2}-0\right)=\underset{h→0}{\lim}\frac{\left[\sin\left(\frac{π}{2}-h\right)+1\right]-\left[\sin\frac{π}{2}+1\right]}{-h}=\underset{h→0}{\lim}\frac{[\cos h+1]-[2]}{-h}=\underset{h→0}{\lim}\frac{1}{h}=∞$