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प्रश्न
If \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, 0 < x < \frac{1}{2}, \text{ find } \frac{dy}{dx} .\] ?
उत्तर
\[\text{ Here, y }= \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}\]
\[\text{ Put 2x }= \cos\theta\]
\[ \therefore y = \cos^{- 1} \left( \cos \theta \right) + 2 \cos^{- 1} \sqrt{1 - \cos^2 \theta}\]
\[ \Rightarrow y = \cos^{- 1} \left( \cos \theta \right) + 2 \cos^{- 1} \left( \sin\theta \right)\]
\[ \Rightarrow y = \cos^{- 1} \left( \cos \theta \right) + 2 \cos^{- 1} \left[ \cos\left( \frac{\pi}{2} - \theta \right) \right] . . . \left( i \right)\]
\[\text{Here}, 0 < x < \frac{1}{2}\]
\[ \Rightarrow 0 < 2x < 1\]
\[ \Rightarrow 0 < \cos\theta < 1\]
\[ \Rightarrow 0 < \theta < \frac{\pi}{2}\]
\[\text{and}\]
\[ \Rightarrow 0 > - \theta > - \frac{\pi}{2}\]
\[ \Rightarrow \frac{\pi}{2} > \left( \frac{\pi}{2} - \theta \right) > 0\]
\[ \Rightarrow 0 < \left( \frac{\pi}{2} - \theta \right) < \frac{\pi}{2}\]
\[\text{ So, from equation} \left( i \right), \]
\[ y = \theta + 2\left( \frac{\pi}{2} - \theta \right) .......\left[ Since, \cos^{- 1} \left( \cos\left( \theta \right) \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right] \right]\]
\[ \Rightarrow y = + \pi - 2\theta\]
\[ \Rightarrow y = \pi - \theta\]
\[ \Rightarrow y = \pi - \cos^{- 1} \left( 2x \right) ........\left[ \text{Since}, 2x = cos\theta \right]\]
Differentiate it with respect to x using chain rule,
\[\frac{d y}{d x} = 0 - \left[ \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right]\frac{d}{dx}\left( 2x \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{\sqrt{1 - 4 x^2}}\left( 2 \right)\]
\[ \therefore \frac{d y}{d x} = \frac{2}{\sqrt{1 - 4 x^2}}\]
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