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