मराठी

If U = Sin − 1 ( 2 X 1 + X 2 ) and V = Tan − 1 ( 2 X 1 − X 2 ) Where − 1 < X < 1 , Then Write the Value of D U D V ? - Mathematics

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प्रश्न

If \[u = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ and v} = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] where \[- 1 < x < 1\], then write the value of \[\frac{du}{dv}\] ?

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उत्तर

\[\text{ We have, u } = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ and }v = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]

\[ \Rightarrow \frac{du}{dx} = \frac{2}{1 + x^2} \text{ and} \frac{dv}{dx} = \frac{2}{1 + x^2} \]

\[ \therefore \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2}{1 + x^2} \times \frac{1 + x^2}{2} = 1\]

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पाठ 11: Differentiation - Exercise 11.09 [पृष्ठ ११८]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.09 | Q 23 | पृष्ठ ११८

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