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प्रश्न
Differentiate \[\cos^{- 1} \left( 4 x^3 - 3x \right)\] with respect to \[\tan^{- 1} \left( \frac{\sqrt{1 - x^2}}{x} \right), \text{ if }\frac{1}{2} < x < 1\] ?
उत्तर
\[\text { Let, u } = \cos^{- 1} \left( 4 x^3 - 3x \right)\]
\[\text { Put, x } = \cos\theta\]
\[ \Rightarrow \theta = \cos^{- 1} x\]
\[\text { Now, u }= \cos^{- 1} \left( 4 \cos^3 \theta - 3\cos\theta \right)\]
\[ \Rightarrow u = \cos^{- 1} \left( \cos3\theta \right) . . . \left( i \right)\]
\[\text { Let, v } = \tan^{- 1} \left( \frac{\sqrt{1 - x^2}}{x} \right)\]
\[ \Rightarrow v = \tan^{- 1} \left( \frac{\sqrt{1 - \cos^2 \theta}}{\cos\theta} \right) \]
\[ \Rightarrow v = \tan^{- 1} \left( \frac{\sin\theta}{\cos\theta} \right)\]
\[ \Rightarrow v = \tan^{- 1} \left( \tan\theta \right) . . . \left( ii \right)\]
\[\text { Here }, \]
\[ \frac{1}{2} < x < 1\]
\[ \Rightarrow \frac{1}{2} < \cos\theta < 1\]
\[ \Rightarrow 0 < \theta < \frac{\pi}{3}\]
\[\text { So, from equation } \left( i \right), \]
\[u = 3\theta .........\left[ \text { Since }, \cos^{- 1} \left( \cos\theta \right) = \theta, \text{ if }\theta \in \left[ 0, \pi \right] \right]\]
\[ \Rightarrow u = 3 \cos^{- 1} x\]
Differentiating it with respect to x,
\[\frac{du}{dx} = \frac{- 3}{\sqrt{1 - x^2}} . . . \left( iii \right)\]
\[\text{ From equation } \left( ii \right), \]
\[v = \theta ..........\left[ \text { Since }, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \right]\]
\[ \Rightarrow v = \cos^{- 1} x\]
Differentiating it with respect to x,
\[\frac{dv}{dx} = \frac{- 1}{\sqrt{1 - x^2}} . . . \left( iv \right)\]
\[\text { Dividing equation} \left( iii \right) \text { by } \left( iv \right), \]
\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \left( \frac{- 3}{\sqrt{1 - x^2}} \right)\left( - \frac{\sqrt{1 - x^2}}{1} \right)\]
\[ \therefore \frac{du}{dv} = 3\]
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