Question

Find the equation of tangent to the curve
$y={\sin }^{-1}\frac{2x}{1+x^2}atx=\sqrt{3}$

Answer 1

$=>y={\sin }^{-1}\frac{2x}{1+x^2}................\left(1\right)$

Let the equation of tangent be $y=mx+c$

$=>m=\left(\frac{dy}{dx}\right)$at$x=\sqrt{3}$

$=>\left(\frac{dy}{dx}\right)=\frac{1}{\sqrt{1-(\frac{2x}{1+x^2}{)}^2}}.\left(\right(2x)\frac{-1\left(2x\right)}{(1+x^2{)}^2}+\frac{2}{(1+{x)}^2})$

$=>\frac{dy}{dx}$ at $x=\sqrt{3}$ $=\frac{1}{\sqrt{1-\frac{4\left(3\right)}{1+3}}}.(\frac{-4\left(3\right)}{({1+3)}^2}+\frac{2}{1+3})$

$=\frac{1}{\sqrt{2}}\frac{-4}{16}$

$=\frac{-1}{4\sqrt{2}}$

Tangent $=>y=\frac{-x}{4\sqrt{2}}+c............\left(2\right)$

at $x=\sqrt{3}$, in $\left(1\right)$, $y={\sin }^{-1}\frac{2\sqrt{3}}{1+3}={\sin }^{-1}\frac{\sqrt{3}}{2}$

$=>y=\frac{\pi }{3}$

Equation $\left(2\right)$ passes through $(\sqrt{3},\frac{\pi }{3})$, $\frac{\pi }{3}=\frac{-\sqrt{3}}{4\sqrt{2}}+c$

So the equation $\left(2\right)$ becomes, $y=\frac{-x}{4\sqrt{2}}+\frac{\pi }{3}+\frac{\sqrt{3}}{4\sqrt{2}}$

$=>\sqrt{2}y=-3x+4\sqrt{2}\pi +3\sqrt{3}.$