# What is the derivative of f(x)=e^(pix)*cos(6x) ?

Sep 7, 2014

The answer is $f ' \left(x\right) = \pi {e}^{\pi x} \cos \left(6 x\right) - 6 {e}^{\pi x} \sin \left(6 x\right)$.

It looks a little complicated, but break it down into pieces that you know how to solve. On the highest level, we see a product of 2 functions, so you should be thinking product rule:

$g \left(x\right) = {e}^{\pi x}$
$h \left(x\right) = \cos \left(6 x\right)$
$f \left(x\right) = g \left(x\right) h \left(x\right)$
and $f ' = g ' \cdot h + g \cdot h '$

Looking at both $g \left(x\right)$ and $h \left(x\right)$, you should notice that both are composition of functions. This means that we need the chain rule:

$g \left(x\right) = j \left(k \left(x\right)\right)$
$g ' \left(x\right) = j ' \left(k \left(x\right)\right) \cdot k ' \left(x\right)$
$g ' \left(x\right) = \pi {e}^{\pi x}$
$h ' \left(x\right) = 6 \sin \left(6 x\right)$

And we get the final answer by substituting:

$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$
$= \pi {e}^{\pi x} \cos \left(6 x\right) + {e}^{\pi x} 6 \sin \left(6 x\right)$

and rearrange to get the answer on the first line.