Differentiate with respect to \(x\)

\[e^{2x}\left( 2 \cos 3x + 3 \sin 3x \right).\]

We’ll start by applying the product rule, defining \(u=e^{2x}\) and \(v=2 \cos 3x + 3 \sin 3x\).

Their derivatives can be calculated via the chain rule, which tells us \(u'=2e^{2x}\).

The function \(v\) contains two trigonometric functions, so we need to recall that \(\dfrac{d (\sin x)}{dx} = \cos x\) and \(\dfrac{d (\cos x)}{dx} = - \sin x\).

Another application of the chain rule then gives \(v' = -6 \sin 3x + 9 \cos x\). We can now combine \(u\), \(v\), \(u'\) and \(v'\) to find

\[\begin{align*} \dfrac{d \left( e^{2x}\left( 2 \cos 3x + 3 \sin 3x \right) \right)}{dx} &= u'v + v'u\\ &= 2e^{2x} \times \left( 2 \cos 3x + 3 \sin 3x \right) + \left( -6 \sin 3x + 9 \cos x \right) \times e^{2x}\\ &= 13 e^{2x} \cos 3x. \end{align*}\]