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प्रश्न
If \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, - \frac{1}{2} < x < 0, \text{ find } \frac{dy}{dx} \] ?
बेरीज
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उत्तर
\[\text{ Here }, y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}\]
\[\text{ Put }2x = \cos\theta\]
\[ \therefore y = \cos^{- 1} \left( \cos \theta \right) + 2 \cos^{- 1} \sqrt{1 - \cos^2 \theta}\]
\[ \Rightarrow y = \cos^{- 1} \left( \cos \theta \right) + 2 \cos^{- 1} \left( \sin\theta \right)\]
\[ \Rightarrow y = \cos^{- 1} \left( \cos \theta \right) + 2 \cos^{- 1} \left[ \cos\left( \frac{\pi}{2} - \theta \right) \right] ........\left( 1 \right)\]
\[\text{Now, }-\frac{1}{2} < x < 0\]
\[ \Rightarrow - 1 < 2x < 0\]
\[ \Rightarrow - 1 < \cos\theta < 0\]
\[ \Rightarrow \frac{\pi}{2} < \theta < \pi\]
And
\[ \Rightarrow - \frac{\pi}{2} > - \theta > - \pi\]
\[ \Rightarrow \left( \frac{\pi}{2} - \frac{\pi}{2} \right) > \left( \frac{\pi}{2} - \theta \right) > \left( \frac{\pi}{2} - \pi \right)\]
\[ \Rightarrow 0 > \left( \frac{\pi}{2} - \theta \right) > - \frac{\pi}{2}\]
\[ \Rightarrow - \frac{\pi}{2} < \left( \frac{\pi}{2} - \theta \right) < 0\]
\[\text{ So, from equation } \left( 1 \right), \]
\[y = \theta + 2\left[ - \left( \frac{\pi}{2} - \theta \right) \right]...........[\text{ Since }, \cos^{- 1} \cos\left( \theta \right) = \theta,\text{ if } \theta \in \left[ 0, \pi \right], \cos^{- 1} \cos\left( \theta \right) = - \theta, \text{ if } \theta \in \left[ - \pi, 0 \right]]\]
\[y = \theta - 2 \times \frac{\pi}{2} + 2\theta\]
\[y = - \pi + 3\theta\]
\[y = - \pi + 3 \cos^{- 1} \left( 2x \right)..........\left[ \text{ Since }, 2x = cos\theta \right]\]
Differentiate it with respect to x using chain rule,
\[\frac{d y}{d x} = 0 + 3\left( \frac{- 1}{\sqrt{1 - \left( 2x \right)^2}} \right)\frac{d}{dx}\left( 2x \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{- 3}{\sqrt{1 - 4 x^2}} \times 2\]
\[ \therefore \frac{d y}{d x} = - \frac{6}{\sqrt{1 - 4 x^2}}\]
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