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प्रश्न
Differentiate \[\cos^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\] ?
उत्तर
\[\text{ Let, y } = \cos^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\]
\[\text{ Put x } = \text{ a cot}\theta\]
\[ \Rightarrow y = \cos^{- 1} \left\{ \frac{a cot\theta}{\sqrt{a^2 co t^2 \theta + a^2}} \right\}\]
\[ \Rightarrow y = \cos^{- 1} \left\{ \frac{a cot\theta}{\sqrt{a^2 \left( co t^2 \theta + 1 \right)}} \right\}\]
\[ \Rightarrow y = \sin^{- 1} \left( \frac{a cot\theta}{\text {a cosec }\theta} \right)\]
\[ \Rightarrow y = \cos^{- 1} \left( \frac{\frac{\cos\theta}{\sin\theta}}{\frac{1}{\sin\theta}} \right)\]
\[ \Rightarrow y = \cos^{- 1} \left( \cos\theta \right) \]
\[ \Rightarrow y = \theta\]
\[ \Rightarrow y = co t^{- 1} \left( \frac{x}{a} \right) \left[ \text{since, x = a cot}\theta \right] \]
\[\text{ Differentiating it with respect to x using chain rule }, \]
\[\frac{d y}{d x} = \frac{- 1}{1 + \left( \frac{x}{a} \right)^2}\frac{d}{dx}\left( \frac{x}{a} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{- a^2}{a^2 + x^2} \times \left( \frac{1}{a} \right)\]
\[ \therefore \frac{d y}{d x} = \frac{- a}{a^2 + x^2}\]
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