$\begin{aligned} & f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]=\sin ^{-1}\left[\frac{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}{10}\right] \\ & \text { Here }(6)^2+(-8)^2=(10)^2 \\ & \therefore \text { Let } \cos \alpha=\frac{6}{10} \text { and } \sin \alpha=\frac{8}{10} \\ & \therefore f(x)=\sin ^{-1}\left[\sin \left(2^x\right) \cos \alpha-\cos \left(2^x\right) \sin \alpha\right]=\sin ^{-1}\left[\sin \left(2^x-\alpha\right)\right] \\ & \therefore f(x)=2^x-\alpha \Rightarrow f^{\prime}(x)=2^x \log 2\end{aligned}$