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Differentiate ( Tan X ) 1 / X ? - Mathematics

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प्रश्न

Differentiate \[\left( \tan x \right)^{1/x}\] ?

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उत्तर

\[\text{ Let y }= \left( \tan x \right)^\frac{1}{x} . . . \left( i \right)\]
\[\text{ Taking log on both sides,} \]
\[\log y = \log \left( \tan x \right)^\frac{1}{x} \]
\[ \Rightarrow \log y = \frac{1}{x}\log\left( \tan x \right) \]
\[\text{ Differentiating with respect to x}, \]
\[\frac{1}{y}\frac{dy}{dx} = \frac{1}{x}\frac{d}{dx}\left\{ \log\left( \tan x \right) \right\} + \log\left( \tan x \right)\frac{d}{dx}\left( \frac{1}{x} \right) \]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} \times \frac{1}{\tan x}\frac{d}{dx}\left( \tan x \right) + \log\left( \tan x \right)\left( - \frac{1}{x^2} \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{x\tan x}\left( \sec^2 x \right) - \frac{\log\left( \tan x \right)}{x^2}\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ \frac{\sec^2 x}{x\tan x} - \frac{\log\left( \tan x \right)}{x^2} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \left( \tan x \right)^\frac{1}{x} \left[ \frac{\sec^2 x}{x\tan x} - \frac{\log\left( \tan x \right)}{x^2} \right] \left[ \text{ using equation } \left( i \right) \right]\]

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अध्याय 11: Differentiation - Exercise 11.05 [पृष्ठ ८८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.05 | Q 16 | पृष्ठ ८८

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