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प्रश्न
Differentiate \[\left( x \cos x \right)^x + \left( x \sin x \right)^{1/x}\] ?
उत्तर
\[\text{ Let y} = \left( x \cos x \right)^x + \left( x \sin x \right)^\frac{1}{x} \]
\[ \text{ Also, Let u } = \left( x \cos x \right)^x \text{ and v } = \left( x \sin x \right)^\frac{1}{x} \]
\[ \therefore y = u + v\]
\[ \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} . . . \left( i \right)\]
\[\text{ Now, u }= \left( x \cos x \right)^x \]
\[ \Rightarrow \log u = \log \left( x \cos x \right)^x \]
\[ \Rightarrow \log u = x \log\left( x \cos x \right)\]
\[ \Rightarrow \log u = x\left[ \log x + \log \cos x \right]\]
\[ \Rightarrow \log u = x\log x + x\log \cos x\]
Differentiate both sides with respect to x,
\[\frac{1}{u}\frac{du}{dx} = \frac{d}{dx}\left( x \log x \right) + \frac{d}{dx}\left( x \log \cos x \right)\]
\[ \Rightarrow \frac{du}{dx} = u\left[ \left\{ \log x\frac{d}{dx}\left( x \right) + x\frac{d}{dx}\left( \log x \right) \right\} + \left\{ \log \cos x\frac{d}{dx}\left( x \right) + x\frac{d}{dx}\left( \log \cos x \right) \right\} \right]\]
\[ \Rightarrow \frac{du}{dx} = \left( x \cos x \right)^x \left[ \left( \log x\left( 1 \right) + x\left( \frac{1}{x} \right) \right) + \left\{ \log \cos x\left( 1 \right) + x\frac{1}{\cos x}\frac{d}{dx}\left( \cos x \right) \right\} \right]\]
\[ \Rightarrow \frac{du}{dx} = \left( x \cos x \right)^x \left[ \left( \log x + 1 \right) + \left\{ \log \cos x + \frac{x}{\cos x}\left( - \sin x \right) \right\} \right]\]
\[ \Rightarrow \frac{du}{dx} = \left( x \cos x \right)^x \left[ \left( 1 + \log x \right) + \left( \log \cos x - x \tan x \right) \right]\]
\[ \Rightarrow \frac{du}{dx} = \left( x \cos x \right)^x \left[ 1 - x \tan x + \left( \log x + \log \cos x \right) \right]\]
\[ \Rightarrow \frac{du}{dx} = \left( x \cos x \right)^x \left[ 1 - x \tan x + \log\left( x \cos x \right) \right] . . . \left( ii \right)\]
\[\text{ Again, v} = \left( x \sin x \right)^\frac{1}{x} \]
\[ \Rightarrow \log v = \log \left( x \sin x \right)^\frac{1}{x} \]
\[ \Rightarrow \log v = \frac{1}{x}\log\left( x \sin x \right)\]
\[ \Rightarrow \log v = \frac{1}{x}\left( \log x + \log \sin x \right)\]
\[ \Rightarrow \log v = \frac{1}{x}\log x + \frac{1}{x}\log \sin x\]
Differentiating both sides with respect to x,
\[\frac{1}{v}\frac{dv}{dx} = \frac{d}{dx}\left( \frac{1}{x}\log x \right) + \frac{d}{dx}\left[ \frac{1}{x}\log\left( \sin x \right) \right]\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = \left[ \log x\frac{d}{dx}\left( \frac{1}{x} \right) + \frac{1}{x}\frac{d}{dx}\left( \log x \right) \right] + \left[ \log\left( \sin x \right)\frac{d}{dx}\left( \frac{1}{x} \right) + \frac{1}{x}\frac{d}{dx}\left\{ \log\left( \sin x \right) \right\} \right]\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = \left[ \log x\left( - \frac{1}{x^2} \right) + \left( \frac{1}{x} \right)\left( \frac{1}{x} \right) \right] + \left[ \log\left( \sin x \right)\left( - \frac{1}{x^2} \right) + \frac{1}{x}\left( \frac{1}{\sin x} \right)\frac{d}{dx}\left( \sin x \right) \right]\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = \frac{1}{x^2}\left( 1 - \log x \right) + \left[ - \frac{\log\left( \sin x \right)}{x^2} + \frac{1}{x \sin x}\left( \cos x \right) \right]\]
\[ \Rightarrow \frac{dv}{dx} = \left( x \sin x \right)^\frac{1}{x} \left[ \frac{1 - \log x}{x^2} + \frac{- \log\left( \sin x \right) + x \cot x}{x^2} \right]\]
\[ \Rightarrow \frac{dv}{dx} = \left( x \sin x \right)^\frac{1}{x} \left[ \frac{1 - \log x - \log\left( \sin x \right) + x \cot x}{x^2} \right]\]
\[ \Rightarrow \frac{dv}{dx} = \left( x \sin x \right)^\frac{1}{x} \left[ \frac{1 - \log\left( x\sin x \right) + x \cot x}{x^2} \right] . . . \left( iii \right)\]
\[\text{ From }\left( i \right), \left( ii \right)\text{ and } \left( iii \right), \text{ we obtain }\]
\[\frac{dy}{dx} = \left( x \cos x \right)^x \left[ 1 - x \tan x + \log\left( x \cos x \right) \right] + \left( x \sin x \right)^\frac{1}{x} \left[ \frac{x \cot x + 1 - \log\left( x \sin x \right)}{x^2} \right]\]
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