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

Answer

`x+y=1`

Explanation

Given the equation of the curve is.
$$ \begin{aligned} &y=(1+x)^{y}+\sin ^{-1}\left(\sin ^{2} x\right) \ldots(1)\\ &y=e^{ \log (1+x)^{y}}+\sin ^{-1}\left(\sin ^{2} x\right) \\ &y=e^{4 \log (1+x)}+\sin ^{-1}\left(\sin ^{2} x\right) \end{aligned} $$
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