# How do you differentiate f(x)=e^x*sin3x using the product rule?

May 12, 2018

The answer is $= {e}^{x} \left(\sin \left(3 x\right) + 3 \cos \left(3 x\right)\right)$

#### Explanation:

The product rule is

$\left(u v\right) ' = u ' v + u v '$

Here,

$f \left(x\right) = {e}^{x} \sin \left(3 x\right)$

and

$u = {e}^{x}$, $\implies$, $u ' = {e}^{x}$

$v = \sin \left(3 x\right)$, $\implies$, $v ' = 3 \cos \left(3 x\right)$

Therefore,

$f ' \left(x\right) = {e}^{x} \sin \left(3 x\right) + 3 {e}^{x} \cos \left(3 x\right)$

$= {e}^{x} \left(\sin \left(3 x\right) + 3 \cos \left(3 x\right)\right)$

May 12, 2018

${e}^{x} \left(\sin 3 x + 3 \cos 3 x\right)$

#### Explanation:

We know,

If $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$,

Then, $f ' \left(x\right) = \frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \left(g \left(x\right) \cdot h \left(x\right)\right) = \frac{d}{\mathrm{dx}} g \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot \frac{d}{\mathrm{dx}} h \left(x\right)$.

This is the Product Rule for Differentiation.

Now,

$f ' \left(x\right)$

$= \frac{d}{\mathrm{dx}} f \left(x\right)$

$= \frac{d}{\mathrm{dx}} \left({e}^{x} \cdot \sin 3 x\right)$

$= \frac{d}{\mathrm{dx}} {e}^{x} \cdot \sin 3 x + \frac{d}{\mathrm{dx}} \left(\sin 3 x\right) \cdot {e}^{x}$

$= {e}^{x} \sin 3 x + {e}^{x} \cos 3 x \cdot \frac{d}{\mathrm{dx}} \left(3 x\right)$ [Chain Rule]

$= {e}^{x} \left(\sin 3 x + 3 \cos 3 x\right)$

Hope this helps.