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

Solution :
$$\eqalign{ & h\left( x \right) = g\left\{ {f\left( x \right)} \right\} = \left[ {f\left( x \right) + 1} \right] = \left[ {\sin \,x + 1} \right] \cr & h'\left( {\frac{\pi }{2} + 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sin \left( {\frac{\pi }{2} + h} \right) + 1} \right] - \left[ {\sin \frac{\pi }{2} + 1} \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\cos \,h + 1} \right] - \left[ 2 \right]}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 2}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - 1}}{h} = - \infty \cr & h'\left( {\frac{\pi }{2} - 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sin \left( {\frac{\pi }{2} - h} \right) + 1} \right] - \left[ {\sin \frac{\pi }{2} + 1} \right]}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\cos \,h + 1} \right] - \left[ 2 \right]}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{1}{h} = + \infty \cr} $$