Single Variable Calculus, Chapter 7, 7.4-2, Section 7.4-2, Problem 34

Differentiate $y = \log_2 (e^{-x} \cos \pi x)$


$
\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} \log_2 (e^{-x} \cos \pi x)
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y' =& \frac{1}{e^{-x} \cos \pi x} \cdot \frac{d}{dx} (e^{-x} \cos \pi x)
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y' =& \frac{1}{e^{-x} \cos \pi x} \left[ (e^{-x}) \frac{d}{dx } (\cos \pi x) + (\cos \pi x) \frac{d}{dx} (e^{-x}) \right]
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y' =& \frac{1}{e^{-x} \cos \pi x} \left[ e^{-x} (- \sin \pi x) \frac{d}{dx} (\pi x) + \cos \pi x (e^{-x}) \right]
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y' =& \frac{1}{e^{-x} \cos \pi x} [-e^{-x} \sin \pi x (\pi) - e^{-x} \cos \pi x]
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y' =& \frac{\cancel{e^{-x}} (- \pi \sin \pi x - \cos \pi x) }{\cancel{e^{-x}} \cos \pi x}
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y' =& \frac{- \pi \sin \pi x - \cos \pi x}{\cos \pi x}
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y' =& \frac{- \pi \sin \pi x}{\cos \pi x} - \frac{\cancel{\cos \pi x}
}{\cancel{\cos \pi x}}
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y' =& - \pi \tan \pi x - 1

\end{aligned}
\end{equation}
$