Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 22

We are given y=log2(exp(-x)*cos(pi*x))=1/ln(2)*ln(exp(-x)*cos(pi*x)). Since the domain of ln(X) is X>0, it means cos(pi*x) cannot equal to zero, we will use this later.
First we use the Chain rule, to the first portion of the answer,
y' = 1/ln(2) * 1/(exp(-x)*cos(pi*x)) * ...
since (ln(f(x)))' = 1/f(x) * f'(x).
The f(x) = exp(-x)*cos(pi*x)
Here use the Product rule, since we have two functions, which are easy to differentiate.
Thus, f'(x) = -exp(-x)*cos(pi*x) + exp(-x)*(-sin(pi*x)*pi)
= -exp(-x) * (cos(pi*x) + sin(pi*x) * pi)
So now y' = 1/ln(2) * 1/(exp(-x)*cos(pi*x)) * -exp(-x) * (cos(pi*x) + sin(pi*x) * pi).
We can immediately cancel the exp(-x). And since cos(pi*x) is never zero, we may divide by it.
Thus, y'(x) = -1/ln(2)*(1 + pi*sin(pi*x)/cos(pi*x))
= -1/ln(2)*(1 + pi*tan(pi*x))