Question

If $y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right),\mathrm{then}\frac{dy}{dx}=$

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

(b) $\frac{2}{1+x^2}$

(c) $\frac{1}{2-x^2}$

(d) $\frac{2}{2-x^2}$

Answer 1

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

$$\begin{gathered} \mathrm{Let}y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\ \mathrm{Differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}x\mathrm{using}\mathrm{chain}\mathrm{rule},\mathrm{we}\mathrm{get}, \\ \frac{dy}{dx}=\frac{1}{\sqrt{1-{\left({\displaystyle \frac{1-x^2}{1+x^2}}\right)}^2}}\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{\left(1+x^2\right)}{\sqrt{{\left(1+x^2\right)}^2-{\left(1-x^2\right)}^2}}\left[\frac{\left(1+x^2\right){\displaystyle \frac{d}{dx}}\left(1-x^2\right)-\left(1-x^2\right){\displaystyle \frac{d}{dx}}\left(1+x^2\right)}{{\left(1+x^2\right)}^2}\right]\left[\mathrm{using}\mathrm{quotient}\mathrm{rule}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{\left(1+x^2\right)}{\sqrt{{\left(1+x^2\right)}^2-{\left(1-x^2\right)}^2}}\left[\frac{\left(1+x^2\right)\left(-2x\right)-\left(1-x^2\right)\left(2x\right)}{{\left(1+x^2\right)}^2}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{\left(1+x^2\right)}{2x}\left[\frac{-2x-2x^3-2x+2x^3}{{\left(1+x^2\right)}^2}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{-4x}{2x\left(1+x^2\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{-2}{1+x^2} \end{gathered}$$