Differentiate the Function: $y=\log_2(e^{-x} \cos(\pi x))$ Differentiate the Function : $y=\log_2(e^{-x} \cos(\pi x))$
Here is my work. What I have I done wrong? 

 A: Your result seems to be okay, you could and should however simplify it. Also, your calculation would have been simpler if you had said something like
$$y=\log_2 e^{-x} \cos\pi x \\
= \frac{-x + \log\cos\pi x}{\log 2} $$
and thus
$$y' \log 2 = -1-\pi\tan\pi x$$
A: $$\frac{-\sin(\pi x)\cdot \pi+\cos(\pi x)\cdot (-1)}{e^x(e^{-x}\cos(\pi x))\cdot \ln 2}$$
is correct, but the last expression you wrote is not correct. You have a mistake when you simplify the expression (you miss minus signs. Also, note that $e^x\cdot e^{-x}=1$). The correct answer will be
$$\frac{-\pi\sin(\pi x)-\cos(\pi x)}{\cos(\pi x)\cdot\ln 2}$$
A: As to the OP's result: just a little cleaning up provides the correct result.

By using 
\begin{align}
\log_{a}(x) = \frac{\log_{b}(x)}{\log_{b}(a)}
\end{align}
then it can be obtained that, using $\log_{e}(x) = \ln(x)$,
\begin{align}
\log_{2}(e^{-x} \, \cos(ax)) = \frac{\ln(e^{-x} \, \cos(ax)}{\ln(2)} = \frac{-x + \ln(\cos(ax))}{\ln(2)}
\end{align}
Now differentiation leads to
\begin{align}
\frac{d}{dx} \left[ \log_{2}(e^{-x} \, \cos(ax)) \right] &= - \frac{1 + a \, \tan(ax)}{\ln(2)} 
\end{align}
