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Q.
The equation of normal to the curve $y=(1+x)^{y}+\sin ^{-1}\left(\sin ^{2}\, x\right)$ at $x=0$ is
Application of Derivatives
Solution:
At $x=0, y=1$.
Hence, the point at which normal is drawn is $P(0,1)$.
Differentiating the given equation w.r.t. $x$, we have
$(1+x)^{y}\{\left.\log (1+x) \frac{d y}{d x}+\frac{y}{1+x}\right\} $
$-\frac{d y}{d x}+\frac{1}{\sqrt{1-\sin ^{4} x}} 2 \sin x \cos x=0$
or $\left(\frac{d y}{d x}\right)_{(0,1)}=\frac{(1+0)^{1} \times \frac{1}{1+0}-\frac{2 \sin 0}{\sqrt{1-\sin ^{2} 0}}}{1-(1+0)^{1} \log 1}=1$
$\therefore \quad$ Slope of the normal $=-1$
Therefore, equation of the normal having slope $-1$ at point $P(0,1)$ is given by
$y-1=-(x-0) $ or $ x+y=1$