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Solution for Differentiate 2 X Cos X ( X 2 + 3 ) 2 ? - CBSE (Science) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

Question

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

Solution

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

Differentiate it with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \frac{2^x \cos x}{\left( x^2 + 3 \right)^2} \right]$

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

$= \left[ \frac{\left( x^2 + 3 \right)^2 \left\{ 2^x \frac{d}{dx}\cos x + \cos x\frac{d}{dx} 2^x \right\} - \left( 2^x \cos x \right)2\left( x^2 + 3 \right)\frac{d}{dx}\left( x^2 + 3 \right)}{\left( x^2 + 3 \right)^4} \right] \left[ \text{Using Product rule and chain rule }\right]$

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

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

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

$So, \frac{d}{dx}\left[ \frac{2^x \cos x}{\left( x^2 + 3 \right)^2} \right] = \frac{2^x}{\left( x^2 + 3 \right)^2}\left[ \cos x \log_e 2 - \sin x - \frac{4x \cos x}{\left( x^2 + 3 \right)} \right]$

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Solution for question: Differentiate 2 X Cos X ( X 2 + 3 ) 2 ? concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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