Question

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

• A. $x-y+1=0$
• B. $x+y+1=0$
• C. $2x-y+1=0$
• D. $x+2y+2=0$

Answer 1

Answer: (A)

$x-y+1=0$

Answer 2

$y={{(1+x)}^{y}}+{{\sin }^{-1}}({{\sin }^{2}}x)$ At $x=0,y=1$ Let $y=u+v$ where $u={{(1+x)}^{y}},$ $v={{\sin }^{-1}}({{\sin }^{2}}x)$ $\Rightarrow$ $\log u=y\log (1+x)$ $\Rightarrow$ $\frac{dy}{dx}\frac{1}{u}=\frac{y}{1+x}+\log (1+x)\frac{dy}{dx}$ and $\frac{dv}{dx}=\frac{1}{\sqrt{1-{{\sin }^{2}}x}}.2\sin x\cos x$ $\therefore$ $\frac{dy}{dx}={{(1+x)}^{y}}\left[ \frac{y}{1+x}+\log (1+x)\frac{dy}{dx} \right]$ $+\frac{\sin 2x}{\cos x}$ $\frac{dy}{dx}[1-{{(1+x)}^{y}}\log (x+1)]$ $=\frac{{{(1+x)}^{y}}y}{1+x}+\frac{\sin 2x}{\cos x}$ $\therefore$ ${{\left. \frac{dy}{dx} \right|}_{(0,1)}}=\frac{1}{1+0}=1$ $\therefore$ Equation of tangent is $(y-1)=1(x)\Rightarrow y-x-1=0$ $\Rightarrow$ $x-y+1=0$