## Calculus 8th Edition

$y=2x+1$
The slope of the tangent can be calculated by taking first derivative of the function $y=e^{2x}cos\pi x$ $y'=e^{2x}\frac{d}{dx}(cos\pi x)+cos\pi x\frac{d}{dx}(e^{2x})$ $y'=2cos\pi x(e^{2x})-\pi (e^{2x})sin\pi x$ Slope of the tangent at (0,1) is given as follows: $y'|_{(0,1)}=2$ The equation of tangent at (0, 1) is $(y-1)=2(x-0)$ Hence, $y=2x+1$