Answer
$$y'=me^{mx}\cos \left(nx\right)-ne^{mx}\sin \left(nx\right)$$
Work Step by Step
Given $$y=e^{mx}cos\left(nx\right)$$
Then
\begin{align*}
y'&=\frac{d}{dx}\left(e^{mx}\right)\cos \left(nx\right)+\frac{d}{dx}\left(\cos \left(nx\right)\right)e^{mx}\\
&=me^{mx}\cos \left(nx\right)-ne^{mx}\sin \left(nx\right)
\end{align*}
