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प्रश्न
Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\cot x} + \left( \cot x \right)^{\tan x}\] ?
उत्तर
\[\text{ We have }, y = \left( \tan x \right)^{\cot x} + \left( \cot x \right)^{\tan x} \]
\[ \Rightarrow y = e^{\log \left( \tan x \right)^{\cot x }}+ e^{\log \left( \cot x \right)^{\tan x}} \]
\[ \Rightarrow y = e^{\cot\ x\ log\ tan\ x}+e^{\tan x \log\left( \cot x \right)} \]
Differentiating with respect to x using chain rule and product rule,
\[\frac{dy}{dx} = \frac{d}{dx}\left( e^{\cot x \log\tan x} \right) + \frac{d}{dx}\left( e^{\tan x logcotx} \right)\]
\[ = e^{\cot x \log\tan x} \frac{d}{dx}\left( {}^{\cot x \log\tan x} \right) + e^{\tan\ x\ logcot x} \frac{d}{dx}\left( {}^{\tan\ x\ logcot\ x} \right)\]
\[ = e^{\log \left( \tan x \right)^{\cot x}}\left[ \cot x\frac{d}{dx}\left( \log \tan x \right) + \log \tan x\frac{d}{dx}\left( \cot x \right) \right] + e^{\log\left( \cot x \right)\tan x} \left[ \tan x\frac{d}{dx}\left( \log \cot x \right) + logcot x\frac{d}{dx}\left( \tan x \right) \right] \]
\[ = \left( \tan x \right)^{\cot x} \left[ \cot x \times \left( \frac{1}{\tan x} \right)\frac{d}{dx}\left( \tan x \right) + \log \tan x\left( - {cosec}^2 x \right) \right] + \left( \cot x \right)^{\tan x} \left[ \tan x \times \left( \frac{1}{\cot x} \right)\frac{d}{dx}\left( \cot x \right) + \log \cot x\left( \sec^2 x \right) \right]\]
\[ = \left( \tan x \right)^{\cot x} \left[ \left( \frac{{cosec}^2 x}{\sec^2 x} \right)\left( \sec^2 x \right) - {cosec}^2 x \log \tan x \right] + \left( \cot x \right)^{\tan x} \left[ \left( \frac{\sec^2 x}{{cosec}^2 x} \right)\left( - {cosec}^2 x \right) + \sec^2 x \log \cot x \right]\]
\[ = \left( \tan x \right)^{\cot x} \left[ {cosec}^2 x - {cosec}^2 x \log \tan x \right] + \left( cot x \right)^{\tan x} \left[ \sec^2 x \log \cot x - \sec^2 x \right]\]
\[ = \left( \tan x \right)^{\cot x} {cosec}^2 x\left[ 1 - \log \tan x \right] + \left( cot x \right)^{\tan x} \sec^2 x \left[ \log \cot x - 1 \right]\]
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