Review question

# Can we differentiate $e^{2x}( 2 \cos (3x) + 3 \sin (3x))$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6530

## Solution

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*}