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Differentiate E X Sin X ( X 2 + 2 ) 3 ? - Mathematics

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प्रश्न

Differentiate \[\frac{e^x \sin x}{\left( x^2 + 2 \right)^3}\] ?

उत्तर

\[\text{Let } y = \frac{e^x \sin x}{\left( x^2 + 2 \right)^3}\]

Differentiate it with respect to  we get,

\[\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^3 \frac{d}{dx}\left( e^x \sin x \right) - e^x \sin x\frac{d}{dx} \left( x^2 + 2 \right)^3}{\left[ \left( x^2 + 2 \right)^3 \right]^2} \left[ \text{Using quotient rule} \right]\]

\[ = \frac{\left( x^2 + 2 \right)^3 \left[ e^x \cos x + \sin x e^x \right] - e^x \sin x 3 \left( x^2 + 2 \right)^2 \left( 2x \right)}{\left( x^2 + 2 \right)^6} \left[ \text{Using product rule} \right]\]

\[ = \frac{\left( x^2 + 2 \right)^3 \left[ e^x \cos x + e^x \sin x \right] - 6x e^x \sin x \left( x^2 + 2 \right)^2}{\left( x^2 + 2 \right)^6}\]

\[ = \frac{\left( x^2 + 2 \right)^2 \left[ \left( x^2 + 2 \right)\left( e^x \cos x + e^x \sin x \right) - 6x e^x \sin x \right]}{\left( x^2 + 2 \right)^6}\]

\[ = \frac{\left( x^2 + 2 \right)\left( e^x \cos x + e^x \sin x \right) - 6x e^x \sin x}{\left( x^2 + 2 \right)^4}\]

\[ = \frac{e^x \sin x + e^x \cos x}{\left( x^2 + 2 \right)^3} - \frac{6x e^x \sin x}{\left( x^2 + 2 \right)^4}\]

\[So, \frac{d y}{d x} = \frac{e^x \sin x + e^x \cos x}{\left( x^2 + 2 \right)^3} - \frac{6x e^x \sin x}{\left( x^2 + 2 \right)^4}\]

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पाठ 11: Differentiation - Exercise 11.02 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.02 | Q 49 | पृष्ठ ३७

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