Question
If $y =\sin ^{-1}\left(2^x\right)$, find $\frac{ d y}{d x}$

Answer

$ y =\sin ^{-1}\left(2^x\right)$
$\therefore \frac{ d y}{ d x}=\frac{ d }{ d x}\left[\sin ^{-1}\left(2^x\right)\right]$
$=\frac{1}{\sqrt{1-\left(2^x\right)^2}} \cdot \frac{ d }{ d x}\left(2^x\right)$
$=\frac{2^x \log 2}{\sqrt{1-\left(2^x\right)^2}}$

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