Question

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

Answer 1

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

Differentiate it with respect to x we get,

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

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

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

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

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

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

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