$$y = \log_7(e^{-x}\cos\pi x)$$
I got: $$y' = \frac {-\sin\pi xe^{-x} - \cos\pi x e^{-x}}{e^{-x}\cos\pi x\ln(7)}$$
Is that correct?
$$y = \log_7(e^{-x}\cos\pi x)$$
I got: $$y' = \frac {-\sin\pi xe^{-x} - \cos\pi x e^{-x}}{e^{-x}\cos\pi x\ln(7)}$$
Is that correct?
In general, the derivative of $\log_a y$, where $y$ is a function of $x$, is
$$\frac{y'}{y \ln a},$$
which I believe you know.
You dropped a $\pi$ when you took the derivative of $\cos (\pi x)$. The derivative is $-\pi \sin(\pi x)$ by the chain rule. Other than that, looks good.