So I differentiated the expression $$ y=e^{2x}\cos(\pi x) $$ And I got $$ y'=2e^{2x}\cos(\pi x)-\pi e^{2x}\sin(\pi x)$$ But when from the given point $(0,1)$ when I plug in zero I get 2
I looked the answer up in the back of my textbook (which I know I shouldn't do) and the answer is $y=2x+1$
You are looking for a line $(y-y_0)=m(x-x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope of the line.
You are correct in finding the derivative, since $m=y'(0)$. That is, the derivative at $x_0$ is the slope of the tangent line at $x_0$. So now you just have to use the point given to find the equation of the line,
$$ (y-1) = 2(x-0) \implies y=2x+1 $$