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

##### 1 Answer

Sep 7, 2014

The answer is

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(x)=e^(pi x)#

#h(x)=cos(6x)#

#f(x)=g(x)h(x)#

and#f'=g'*h+g*h'#

Looking at both

#g(x)=j(k(x))#

#g'(x)=j'(k(x))*k'(x)#

#g'(x)=pi e^(pi x)#

#h'(x)=6 sin(6x)#

And we get the final answer by substituting:

#f'(x)=g'(x)h(x)+g(x)h'(x)#

#=pi e^(pi x) cos(6x)+ e^(pi x)6 sin(6x)#

and rearrange to get the answer on the first line.