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प्रश्न
Differentiate \[\cos^{- 1} \left( \frac{1 - x^{2n}}{1 + x^{2n}} \right), < x < \infty\] ?
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उत्तर
\[\text{Let, y } = \cos^{- 1} \left( \frac{1 - x^{2n}}{1 + x^{2n}} \right)\]
\[\text{ Put } x^n = \tan\theta\]
\[ \therefore y = \cos^{- 1} \left[ \frac{1 - \left( x^n \right)^2}{1 + \left( x^n \right)^2} \right]\]
\[ \Rightarrow y = \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \]
\[ \Rightarrow y = \cos^{- 1} \left( \cos 2\theta \right) . . . \left( i \right)\]
\[\text{ Here }, 0 < x < \infty \]
\[ \Rightarrow 0 < x^n < \infty \]
\[ \Rightarrow 0 < \tan\theta < \infty \]
\[ \Rightarrow 0 < \theta < \frac{\pi}{2}\]
\[ \Rightarrow 0 < 2\theta < \pi\]
\[\text{ So, from equation } \left( i \right), \]
\[ y = 2\theta \left[ \text{ Since }, \cos^{- 1} \left( \cos \theta \right) = \theta, if \theta \in \left[ 0, \pi \right] \right]\]
\[ \Rightarrow y = 2 \tan^{- 1} \left( x^n \right)\]
Differentiate it with respect to x using chain rule,
\[\frac{d y}{d x} = 2\left[ \frac{1}{1 + \left( x^n \right)^2} \right]\frac{d}{dx}\left( x^n \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{2}{1 + x^{2n}} \times \left( n x^{n - 1} \right)\]
\[ \therefore \frac{d y}{d x} = \frac{2n x^{n - 1}}{1 + x^{2n}}\]
