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प्रश्न
Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\log x} + \cos^2 \left( \frac{\pi}{4} \right)\] ?
उत्तर
\[\text{ We have, y }= \left( \tan x \right)^{\log x }+ \cos^2 \left( \frac{\pi}{4} \right)\]
\[ \Rightarrow y = e^{ \log \left( \tan x \right)^{\log x }} + \cos^2 \left( \frac{\pi}{4} \right)\]
\[ \Rightarrow y = e^{ \log x \log \tan x }+ \cos^2 \left( \frac{\pi}{4} \right)\]
Differentiating with respect to x using chain rule,
\[\frac{dy}{dx} = \frac{d}{dx}\left( e^{\log x \log \tan x} \right) + \frac{d}{dx} \cos^2 \left( \frac{\pi}{4} \right)\]
\[ = e^{\log x \log \tan x } \frac{d}{dx}\left( \log x \log \tan x \right) + 0\]
\[ = e^{\log \left( \tan x \right)^{\log x}} \left[ \log x\frac{d}{dx}\left( \log \tan x \right) + \log \tan x\frac{d}{dx}\left( \log x \right) \right] \]
\[ = \left( \tan x \right)^{\log x } \left[ \log x\left( \frac{1}{\tan x} \right)\frac{d}{dx}\left( \tan x \right) + \log \tan x\left( \frac{1}{x} \right) \right]\]
\[ = \left( \tan x \right)^{\log x} \left[ \log x\left( \frac{1}{\tan x} \right)\left( \sec^2 x \right) + \frac{\log \tan x}{x} \right]\]
\[ = \left( \tan x \right)^{\log x } \left[ \log x\left( \frac{\sec^2 x}{\tan x} \right) + \frac{\log \tan x}{x} \right]\]
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