मराठी

Differentiate Tan − 1 { X 1 / 3 + a 1 / 3 1 − ( a X ) 1 / 3 } ?

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प्रश्न

Differentiate \[\tan^{- 1} \left\{ \frac{x^{1/3} + a^{1/3}}{1 - \left( a x \right)^{1/3}} \right\}\] ?

बेरीज
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उत्तर

\[\text{ Let, y } = \tan^{- 1} \left[ \frac{x^\frac{1}{3} + a^\frac{1}{3}}{1 - \left( ax \right)^\frac{1}{3}} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left( x^\frac{1}{3} \right) + \tan^{- 1} \left( a^\frac{1}{3} \right) \left[ \text{ Since }, \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right) \right]\]

Differentiate it with respect to x using chain rule,

\[\frac{d y}{d x} = \frac{1}{1 + \left( x^\frac{1}{3} \right)^2} \times \frac{d}{dx}\left( x^\frac{1}{3} \right) + 0\]

\[ \Rightarrow \frac{d y}{d x} = \frac{\left( \frac{1}{3} \times x^{\frac{1}{3} - 1} \right)}{1 + x^\frac{2}{3}}\]

\[ \therefore \frac{d y}{d x} = \frac{1}{3 x^\frac{2}{3} \left( 1 + x^\frac{2}{3} \right)}\]

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पाठ 10: Differentiation - Exercise 11.03 [पृष्ठ ६४]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 10 Differentiation
Exercise 11.03 | Q 33 | पृष्ठ ६४
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