Find the equation of the normal to the curve `y=(1+x)^y+sin^(-1)(sin^2x)` at `x=0.`

Answer

Equation of the normal to the given curve (1) at `(0,1)` is,
`x+y=1`

Explanation

Equation of the curves are:
$$ \begin{aligned} &y=(1+x)^{y}+\sin ^{-1}(\sin 2 x)-(1) \\ &y=e^{\log (1+x) y}+\sin ^{-1}\left(\sin ^{2} x\right) \\ &y=e^{y \log (1+x)}+\sin ^{-1}\left(\sin ^{2} x\right) \end{aligned} $$
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