Question

If \[y = \sin^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right), \text { then } \frac{dy}{dx} =\] _____________ .

पर्याय

• \[- \frac{2}{1 + x^2}\]

• \[\frac{2}{1 + x^2}\]

• \[\frac{1}{2 - x^2}\]

• \[\frac{2}{2 - x^2}\]

Answer 1

\[- \frac{2}{1 + x^2}\]

 

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