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If Y = Log √ 1 + Tan X 1 − Tan X Prove that D Y D X = Sec 2 X ? - Mathematics

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प्रश्न

If  \[y = \log \sqrt{\frac{1 + \tan x}{1 - \tan x}}\]  prove that \[\frac{dy}{dx} = \sec 2x\] ?

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उत्तर

\[\text{Let y } = \log\sqrt{\frac{1 + \tan x}{1 - \tan x}}\]

\[ \Rightarrow y = \log \left\{ \frac{1 + \tan x}{1 - \tan x} \right\}^\frac{1}{2} \]

\[ \Rightarrow y = \frac{1}{2}\log\left\{ \frac{1 + \tan x}{1 - \tan x} \right\}\]

\[ \Rightarrow y = \frac{1}{2}\left\{ \log\left( 1 + \tan x \right) - \log\left( 1 - \tan x \right) \right\}\]

\[ \Rightarrow \frac{d y}{d x} = \frac{1}{2}\left\{ \frac{d}{dx}\left\{ \log\left( 1 + \tan x \right) \right\} - \frac{d}{dx}\left\{ \log\left( 1 - \tan x \right) \right\} \right\}\]

\[ = \frac{1}{2}\left\{ \frac{1}{1 + \tan x } \times \frac{d}{dx}\left( 1 + \tan x \right) - \frac{1}{1 - \tan x} \times \frac{d}{dx}\left( 1 - \tan x \right) \right\}\]

\[ = \frac{1}{2}\left\{ \frac{1}{1 + \tan x}\left( 0 + \sec^2 x \right) - \frac{1}{1 - \tan x}\left( 0 - \sec^2 x \right) \right\}\]

\[ = \frac{1}{2}\left\{ \frac{\sec^2 x}{1 + \tan x} + \frac{\sec^2 x}{1 - \tan x} \right\}\]

\[ = \frac{1}{2} \sec^2 x\left\{ \frac{1 - \tan x + 1 + \tan x}{1 - \tan^2 x} \right\}\]

\[ = \frac{1}{2} \sec^2 x\left( \frac{2}{1 - \tan^2 x} \right)\]

\[ = \frac{\sec^2 x}{1 - \tan^2 x}\]

\[ = \frac{1 + \tan^2 x}{1 - \tan^2 x}\]

\[ = \frac{1}{\frac{1 - \tan^2 x}{1 + \tan^2 x}}\]

\[ = \frac{1}{\cos2x}\]

\[ = \sec2x\]

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अध्याय 11: Differentiation - Exercise 11.02 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.02 | Q 62 | पृष्ठ ३८

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