Let  $f\left(x\right)=\underset{h\to 0}{\lim }\left(\frac{\sin {(x+h)}^{1n(x+h)}-\sin x^{1nx}}{h}\right)$  then  $f\left(\frac{\pi }{2}\right)$  is

$$\begin{gathered} Let g\left(x\right)=\sin x^{lnx} \\ {g}^1\left(x\right)=\underset{h\to o}{\lim }\frac{\sin {(x+n)}^{\ln (x+h)}-\sin x^{\ln x}}{h} \\ \therefore f\left(x\right)=g^1\left(x\right)=\sin x^{\ln x}(\frac{\ln \sin x}{x}+\cot x.\ln x) \\ f\left(\frac{\pi }{2}\right)=1(0+0)=0. \end{gathered}$$