PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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Solution for Differentiate E X Sin X ( X 2 + 2 ) 3 ? - PUC Karnataka Science Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

Question

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

Solution

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

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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Solution Differentiate E X Sin X ( X 2 + 2 ) 3 ? Concept: Simple Problems on Applications of Derivatives.
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