Question

Differentiate \[e^{3 x} \cos 2x\] ?

Answer 1

\[\text{ Let } y = e^{3x} \cos2x\]

\[\text{ Differentiate it with respect to x we get }, \]

\[\frac{d y}{d x} = \frac{d}{dx}\left( e^{3x} \cos2x \right)\]

\[ = e^{3x} \times \frac{d}{dx}\left( \cos2x \right) + \cos2x\frac{d}{dx}\left( e^{3x} \right) \left[ \text{Using product rule } \right]\]

\[ = e^{3x} \times \left( - \sin2x \right)\frac{d}{dx}\left( 2x \right) + \cos2x e^{3x} \frac{d}{dx}\left( 3x \right) \left[ \text{ Using chain rule } \right]\]

\[ = - 2 e^{3x} \sin2x + 3 e^{3x} \cos2x\]

\[ = e^{3x} \left( 3 \cos2x - 2 \sin2x \right)\]

\[So, \frac{d}{dx}\left( e^{3x} \cos2x \right) = e^{3x} \left( 3 \cos2x - 2 \sin2x \right)\]