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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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