Question

Find $\frac{dy}{dx} y=sin^{-1}(\frac{1-x^2}{1+x^2}),0<x<1$

Answer 1

The given relationship is y=sin-1(1-x2/1+x2)$y=sin^{-1}(\frac{1-x^2}{1+x^2})$

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

$y=sin-1(\frac{1-x2}{1+x2})$
$⇒siny=(\frac{1-x2}{1+x2})$
$⇒(1+x^2)siny=1-x^2$
$⇒(1+siny)x2^=1-siny$
$⇒x^2=\frac{1-siny}{1+siny}$
$⇒x^2=\frac{(cos\frac{y}{2}-sin\frac{y}{2})}{(cos\frac{y}{2}+sin\frac{y}{2})^2}$
$⇒x=\frac{cos\frac{y}{2}-sin\frac{y}{2}}{cos\frac{y}{2}+sin\frac{y}{2}}$
$⇒x=tan(\frac{π}{4}-\frac{y}{2})$
Differentiating this relationship with respect to x, we obtain
$\frac{d}{dx}(x)=\frac{d}{dx}.[tan(\frac{π}{4}-\frac{y}{2})]$
$⇒\frac{dy}{dx}=\frac{-2}{1+x^2}$