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प्रश्न
Differentiate \[x^{x \cos x +} \frac{x^2 + 1}{x^2 - 1}\] ?
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उत्तर
\[\text{ Let y }= x^{x \cos x} + \frac{x^2 + 1}{x^2 - 1}\]
\[\text{ Also, Let u } = x^{x \cos x} \text{ and v } = \frac{x^2 + 1}{x^2 - 1}\]
\[ \therefore y = u + v\]
\[ \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} . . . \left( i \right)\]
\[\text{ Now, u }= x^{x \cos x} \]
\[ \Rightarrow \log u = \log\left( x^{x \cos x} \right)\]
\[ \Rightarrow \log u = x \cos x \log x\]
Differentiating both sides with respect to x,
\[\frac{1}{u}\frac{du}{dx} = \cos x \log x\frac{d}{dx}\left( x \right) + x\log x\frac{d}{dx}\left( \cos x \right) + x \cos x\frac{d}{dx}\left( \log x \right)\]
\[ \Rightarrow \frac{du}{dx} = u\left[ \cos x \log x + x\left( - \sin x \right)\log x + x \cos x\left( \frac{1}{x} \right) \right]\]
\[ \Rightarrow \frac{du}{dx} = x^{x \cos x} \left( \cos x \log x - x \sin x \log x + \cos x \right)\]
\[ \Rightarrow \frac{du}{dx} = x^{x \cos x} \left[ \cos x\left( 1 + \log x \right) - x \sin x \log x \right] . . . \left( 2 \right)\]
\[\text{ Again, v }= \frac{x^2 + 1}{x^2 - 1}\]
\[ \Rightarrow \log v = \log\left( x^2 + 1 \right) - \log\left( x^2 - 1 \right)\]
Differentiating both sides with respect to x,
\[\frac{1}{v}\frac{dv}{dx} = \frac{2x}{x^2 + 1} - \frac{2x}{x^2 - 1}\]
\[ \Rightarrow \frac{dv}{dx} = v\left[ \frac{2x\left( x^2 - 1 \right) - 2x\left( x^2 + 1 \right)}{\left( x^2 + 1 \right)\left( x^2 - 1 \right)} \right]\]
\[ \Rightarrow \frac{dv}{dx} = \frac{x^2 + 1}{x^2 - 1}\left[ \frac{- 4x}{\left( x^2 + 1 \right)\left( x^2 - 1 \right)} \right]\]
\[ \Rightarrow \frac{dv}{dx} = \frac{- 4x}{\left( x^2 - 1 \right)^2} . . . \left( 3 \right)\]
\[\text{ From} \left( i \right), \left( ii \right) \text{ and } \left( iii \right), \text{ we obtain}\]
\[\frac{dy}{dx} = x^{x \cos x} \left[ \cos x\left( 1 + \log x \right) - x \sin x \log x \right] - \frac{4x}{\left( x^2 - 1 \right)^2}\]
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