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Differentiate Cos − 1 { X √ X 2 + a 2 } ? - Mathematics

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प्रश्न

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

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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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अध्याय 11: Differentiation - Exercise 11.03 [पृष्ठ ६३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.03 | Q 9 | पृष्ठ ६३

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