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

\[\text{ Let, y } = \tan^{- 1} \left( \frac{\sqrt{x} + \sqrt{a}}{1 - \sqrt{xa}} \right)\] \[ \Rightarrow y = \tan^{- 1} \sqrt{x} + \tan^{- 1} \sqrt{a} \left[ \text{ Since }, \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \frac{x + y}{1 - xy} \right]\] Differentiate it with respect to x using chain rule, \[\frac{d y}{d x} = \frac{d}{dx}\left( \tan^{- 1} \sqrt{x} \right) + \frac{d}{dx}\left( \tan^{- 1} \sqrt{a} \right)\] \[ \Rightarrow \frac{d y}{d x} = \frac{1}{1 + \left( \sqrt{x} \right)^2}\frac{d}{dx}\left( \sqrt{x} \right) + 0\] \[ \Rightarrow \frac{d y}{d x} = \left( \frac{1}{1 + x} \right)\left( \frac{1}{2\sqrt{x}} \right)\] \[ \therefore \frac{d y}{d x} = \frac{1}{2\sqrt{x}\left( 1 + x \right)}\]

Differentiate Tan − 1 ( √ X + √ a 1 − √ X a ) ? - Mathematics

प्रश्न

Differentiate $\tan^{- 1} (\frac{\sqrt{x} + \sqrt{a}}{1 - \sqrt{x a}})$ ?

उत्तर

$Let, y = \tan^{- 1} (\frac{\sqrt{x} + \sqrt{a}}{1 - \sqrt{x a}})$

$\Rightarrow y = \tan^{- 1} \sqrt{x} + \tan^{- 1} \sqrt{a} [Since, \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \frac{x + y}{1 - x y}]$

Differentiate it with respect to x using chain rule,

$\frac{d y}{d x} = \frac{d}{d x} (\tan^{- 1} \sqrt{x}) + \frac{d}{d x} (\tan^{- 1} \sqrt{a})$

$\Rightarrow \frac{d y}{d x} = \frac{1}{1 + {(\sqrt{x})}^{2}} \frac{d}{d x} (\sqrt{x}) + 0$

$\Rightarrow \frac{d y}{d x} = (\frac{1}{1 + x}) (\frac{1}{2 \sqrt{x}})$

$∴ \frac{d y}{d x} = \frac{1}{2 \sqrt{x} (1 + x)}$

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Simple Problems on Applications of Derivatives

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Q 26 Q 25 Q 27

अध्याय 11: Differentiation - Exercise 11.03 [पृष्ठ ६३]

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अध्याय 11 Differentiation
Exercise 11.03 | Q 26 | पृष्ठ ६३

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Simple Problems on Applications of Derivatives

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Differentiate Tan − 1 ( √ X + √ a 1 − √ X a ) ? - Mathematics

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