Question

Differentiate \[\sin^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\] ?

Answer 1

\[\text{ Let, y }= \sin^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\]

\[\text{put x } = a \tan\theta\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \frac{a \tan\theta}{\sqrt{a^2 \tan^2 \theta + a^2}} \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left\{ \frac{a \tan\theta}{\sqrt{a^2 \left( \tan^2 \theta + 1 \right)}} \right\}\]

\[ \Rightarrow y = \sin^{- 1} \left( \frac{a \tan\theta}{a sec\theta} \right) \]

\[ \Rightarrow y = \sin^{- 1} \left( \sin\theta \right) \]

\[ \Rightarrow y = \theta\]

\[ \Rightarrow y = \tan^{- 1} \left( \frac{x}{a} \right) \left[ \text{since, }x = a \tan\theta \right] \]

\[\text{ Differentiating it with respect to x using chain rule }, \]

\[ \frac{d y}{d x} = \frac{1}{1 + \left( \frac{x}{a} \right)^2}\frac{d}{dx}\left( \frac{x}{a} \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{a^2}{a^2 + x^2} \times \left( \frac{1}{a} \right)\]

\[ \therefore \frac{d y}{d x} = \frac{a}{a^2 + x^2}\]