Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 24

Determine the function $y = \log_2 (e^{-x}\cos \pi x)$

$
\begin{equation}
\begin{aligned}
y' &= \frac{1}{(e^{-x}\cos \pi x) \ln 2} \cdot \frac{d}{dx} (e^x \cos \pi x)\\
\\
y' &= \frac{1}{(e^{-x}\cos \pi x) \ln 2} \left[ e^{-x} \frac{d}{dx} (\cos \pi x) + (\cos \pi x) \frac{d}{dx} \left( \frac{1}{e^x} \right) \right]\\
\\
y' &= \frac{1}{(e^{-x}\cos \pi x) \ln 2} \left[ e^{-x} (- \sin \pi x) \frac{d}{dx} (\pi x) + (\cos \pi x) \left( \frac{-1}{e^x} \right)\right]\\
\\
y' &= \frac{1}{(e^{-x}\cos \pi x) \ln 2} \left[ -e^{-x} \sin \pi x \cdot \pi - e^x \cos (\pi x) \right]\\
\\
y' &= \frac{-\pi e^{-x} \sin \pi x - e^{-x} \cos \pi x }{e^{-x} \cos \pi x \ln 2}\\
\\
y' &= \frac{- \pi \cancel{e^{-x}} \sin \pi x}{\cancel{e^{-x}} \cos \pi x \ln 2} - \frac{\cancel{e^{-x}} \cancel{\cos \pi x}}{\cancel{e^{-x}} \cancel{\cos \pi x} \ln 2}\\
\\
y' &= \frac{- \pi \tan \pi x}{\ln 2} - \frac{1}{\ln 2}\\
\\
y' &= - \frac{1 + \pi \tan \pi x}{\ln 2}
\end{aligned}
\end{equation}
$