Question

Let $si{n}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{dy}{dx}$ is

• A. $\frac{2}{1+x^2}$
• B. $\frac{1}{2(1+x^2)}$
• C. $\frac{-2}{1+x^2}$
• D. $\frac{2}{2-x^2}$

Answer 1

Answer: (C)

$\frac{-2}{1+x^2}$

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

Differentiating both sides w.r.t $x$

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

$=\frac{1+x^2}{\sqrt{{(1+x^2)}^2-{(1-x^2)}^2}}\left[\frac{(1+x^2)(-2x)-(1-x^2)\left(2x\right)}{{(1+x^2)}^2}\right]$

$=\frac{1}{\sqrt{(1+x^2-1+x^2)(1+x^2-1-x^2)}}\left[\frac{-2x(1+x^2+1-x^2)}{{(1+x^2)}^2}\right]$

$=\frac{-2x\times 2}{\sqrt{2x^2\times 2}(1+x^2)}$

$=\frac{-4x}{2x(1+x^2)}=\frac{-2}{(1+x^2)}$