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प्रश्न
Differentiate \[x^\left( \sin x - \cos x \right) + \frac{x^2 - 1}{x^2 + 1}\] ?
बेरीज
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उत्तर
\[\text{Let y } = x^\left( \sin x - \cos x \right) + \left( \frac{x^2 - 1}{x^2 + 1} \right)\]
\[ \Rightarrow y = e^{\log x^\left( \sin x - \cos x \right)} + \left( \frac{x^2 - 1}{x^2 + 1} \right)\]
\[ \Rightarrow y = e^{ \left( \sin x - \cos x \right)\log x } + \left( \frac{x^2 - 1}{x^2 + 1} \right)\]
Differentiate it with respect to x using chain rule,
\[\frac{dy}{dx} = \frac{d}{dx}\left[ e^{\left( \sin x - \cos x \right)\log x }\right] + \frac{d}{dx}\left[ \frac{x^2 - 1}{x^2 + 1} \right]\]
\[ = e^{\left( \sin x - \cos x \right)\log x }\frac{d}{dx}\left\{ \left( \sin x - \cos x \right)\log x \right\} + \left[ \frac{\left( x^2 + 1 \right)\frac{d}{dx}\left( x^2 - 1 \right) - \left( x^2 - 1 \right)\frac{d}{dx}\left( x^2 + 1 \right)}{\left( x^2 + 1 \right)^2} \right]\]
\[ = e^{ \log x^\left( \sin x - \cos x \right) } \left[ \left( \sin x - \cos x \right)\frac{d}{dx}\left( \log x \right) + \left( \log x \right)\frac{d}{dx}\left( \sin x - \cos x \right) \right] + \left[ \frac{\left( x^2 + 1 \right)\left( 2x \right) - \left( x^2 - 1 \right)\left( 2x \right)}{\left( x^2 + 1 \right)^2} \right]\]
\[ = x^\left( \sin x - \cos x \right) \left[ \left( \sin x - \cos x \right)\left( \frac{1}{x} \right) + \log x\left( \sin x + \cos x \right) \right] + \left[ \frac{2 x^3 + 2x - 2 x^3 + 2x}{\left( x^2 + 1 \right)^2} \right]\]
\[ = x^\left( \sin x - \cos x \right) \left[ \frac{\left( \sin x - \cos x \right)}{x} + \left( \sin x + \cos x \right)\log x \right] + \frac{4x}{\left( x^2 + 1 \right)^2}\]
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