Question

(x sin x + cos x ) (e^x + x² log x) 

Answer 1

\[\text{ Let } u = x \sin x + \cos x; v = e^x + x^2 \log x \]
\[\text{ Then }, u' = \left[ x\frac{d}{dx}\left( \sin x \right) + \sin x \frac{d}{dx}\left( x \right) \right] - \sin x \]
\[ = x \cos x + \sin x - \sin x \]
\[ = x \cos x\]
\[ v' = e^x + \left[ x^2 \frac{d}{dx}\left( \log x \right) + \log x \frac{d}{dx}\left( x^2 \right) \right]\]
\[ = e^x + x + 2x \log x \]
\[ \]
\[ \text{ Using the product rule }:\]
\[\frac{d}{dx}\left( uv \right) = u v ' + v u'\]
\[\frac{d}{dx}\left[ \left( x \sin x + \cos x \right)\left( e^x + x^2 \cos x \right) \right]\]
\[ = \left( x \sin x + \cos x \right)\left( e^x + x + 2x \log x \right) + \left( e^x + x^2 \log x \right) \left( x \cos x \right)\]
\[\]