Friday, July 15, 2016

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}
$

No comments:

Post a Comment

Why is the fact that the Americans are helping the Russians important?

In the late author Tom Clancy’s first novel, The Hunt for Red October, the assistance rendered to the Russians by the United States is impor...