Question

If $y=({\sin }^{-1}x{)}^x$, find $\frac{dy}{dx}$

Answer 1

From question,

$y=(si{n}^{-1}x{)}^x$

Taking log, we get

$logy=x.log\left(si{n}^{-1}x\right)$

On differentiating, w.r.t $x$ we get,

$\frac{1}{y}.\frac{dx}{dy}=log\left(si{n}^{-1}x\right)+x.\frac{1}{si{n}^{-1}x}.\frac{1}{\sqrt{1-x^2}}$

$⟹\frac{dx}{dy}=y.\left[log\right(si{n}^{-1}x)+x.\frac{1}{si{n}^{-1}x}.\frac{1}{\sqrt{1-x^2}}]$

$⟹\frac{dx}{dy}=(si{n}^{-1}x{)}^x.[log\left(si{n}^{-1}x\right)+x.\frac{1}{si{n}^{-1}x}.\frac{1}{\sqrt{1-x^2}}]$