Discussion Forum

The equation of normal to the curve $y=(1+x)^{y}+\sin ^{-1}(\sin^{2}x)$ at x=0 is 

Answer:

Explanation:

Given curve is 

  $y-(1+x)^{y}+ \sin^{-1} (\sin^{2}x)$

 On  differentiating w.r.t x , we get

  $\frac{dy}{dx}=(1+x)^{y}\left[\frac{y}{1+x}+\log (1+x)\frac{dy}{dx}\right]+\frac{2 \sin x \cos x}{\sqrt{1-\sin^{4}x }}$

$\Rightarrow$     $\left(\frac{dy}{dx}\right)_{at(0,1)}=1 [ \because at x=0,y=1]$

 Slope of normal at (x=0)=-1

$\therefore$  Equation of normal at x=0 and y=1 is  y-1=-1(x-0)

$\Rightarrow$    $y-1=-x \Rightarrow x+y=1$