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