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If Y = Sin − 1 ( X 1 + X 2 ) + Cos − 1 ( 1 √ 1 + X 2 ) , 0 < X < ∞ Prove that D Y D X = 2 1 + X 2 ? - Mathematics

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प्रश्न

If \[y = \sin^{- 1} \left( \frac{x}{1 + x^2} \right) + \cos^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right), 0 < x < \infty\] prove that  \[\frac{dy}{dx} = \frac{2}{1 + x^2} \] ?

 

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उत्तर

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

\[\text{ Put x} = \tan\theta\]

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

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

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

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

\[\text{ Here,} 0 < x < \infty \]

\[ \Rightarrow 0 < \tan\theta < \infty \]

\[ \Rightarrow 0 < \theta < \frac{\pi}{2}\]

\[\text{ So, from equation} \left( i \right), \]

\[y = \theta + \theta ...........[\text{Since, }\sin^{- 1} \left( \sin\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right], \cos^{- 1} \left( \cos\theta \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right]\]

\[ \Rightarrow y = 2\theta\]

\[ \Rightarrow y = 2 \tan^{- 1} x ...........\left[ \text{Since}, x = \tan\theta \right]\]

Differentiate it with respect to x,

\[\therefore \frac{d y}{d x} = \frac{2}{1 + x^2}\]

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अध्याय 11: Differentiation - Exercise 11.03 [पृष्ठ ६४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.03 | Q 36 | पृष्ठ ६४

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