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प्रश्न
If \[y = \cos^{- 1} \left\{ \frac{2x - 3 \sqrt{1 - x^2}}{\sqrt{13}} \right\}, \text{ find } \frac{dy}{dx}\] ?
उत्तर
\[\text{ Let y }= \cos^{- 1} \left\{ \frac{2x - 3\sqrt{1 - x^2}}{\sqrt{13}} \right\}\]
\[\text{ Put, x }= \cos\theta\]
\[ y = \cos^{- 1} \left\{ \frac{2\cos\theta - 3\sqrt{1 - \cos^2 \theta}}{\sqrt{13}} \right\}\]
\[ y = \cos^{- 1} \left\{ \frac{2\cos\theta - 3\sin\theta}{\sqrt{13}} \right\}\]
\[ y = \cos^{- 1} \left\{ \cos\theta\left( \frac{2}{\sqrt{13}} \right) + \sin\theta\left( \frac{3}{\sqrt{13}} \right) \right\}\]
\[\text{ Let } \cos\phi = \frac{2}{\sqrt{13}} \text{ and }\sin\phi = \frac{3}{\sqrt{13}}\]
\[ y = \cos^{- 1} \left\{ \cos\theta\cos\phi + \sin\theta \sin\phi \right\}\]
\[ y = \cos^{- 1} \left\{ \cos\left( \theta - \phi \right) \right\} . . . \left( i \right)\]
\[ y = \left( \theta - \phi \right) \]
\[ y = \cos^{- 1} x - \phi \]
\[\text{ Differentiating it with respect to x }, \]
\[\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^2}} + 0\]
\[\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^2}}\]
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