हिंदी

Differentiate Tan − 1 ( a + X 1 − a X ) ?

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प्रश्न

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

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उत्तर

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

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

Differentiate it with respect to x,

\[\frac{d y}{d x} = \frac{d}{dx}\left( \tan^{- 1} a \right) + \frac{d}{dx}\left( \tan^{- 1} x \right)\]

\[ \Rightarrow \frac{d y}{d x} = 0 + \frac{1}{1 + x^2}\]

\[ \therefore \frac{d y}{d x} = \frac{1}{1 + x^2}\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Differentiation - Exercise 11.03 [पृष्ठ ६३]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 10 Differentiation
Exercise 11.03 | Q 25 | पृष्ठ ६३
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