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Question
Differentiate \[\sin^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\] ?
Solution
\[\text{ Let, y }= \sin^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\]
\[\text{put x } = a \tan\theta\]
\[ \Rightarrow y = \sin^{- 1} \left\{ \frac{a \tan\theta}{\sqrt{a^2 \tan^2 \theta + a^2}} \right\}\]
\[ \Rightarrow y = \sin^{- 1} \left\{ \frac{a \tan\theta}{\sqrt{a^2 \left( \tan^2 \theta + 1 \right)}} \right\}\]
\[ \Rightarrow y = \sin^{- 1} \left( \frac{a \tan\theta}{a sec\theta} \right) \]
\[ \Rightarrow y = \sin^{- 1} \left( \sin\theta \right) \]
\[ \Rightarrow y = \theta\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{x}{a} \right) \left[ \text{since, }x = a \tan\theta \right] \]
\[\text{ Differentiating it with respect to x using chain rule }, \]
\[ \frac{d y}{d x} = \frac{1}{1 + \left( \frac{x}{a} \right)^2}\frac{d}{dx}\left( \frac{x}{a} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{a^2}{a^2 + x^2} \times \left( \frac{1}{a} \right)\]
\[ \therefore \frac{d y}{d x} = \frac{a}{a^2 + x^2}\]
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