## Calculus: Early Transcendentals 8th Edition

The equation of the tangent line $(l)$ to $y$ at point $(0,1)$ would be $$(l): y=x+1$$
$$y=e^x\cos x$$ 1) Find the derivative of $y$ $$y'=(e^x\cos x)'$$ $$y'=(e^x)'\cos x+e^x(\cos x)'$$ $$y'=e^x\cos x-e^x\sin x$$ 2) Find $y'(0)$ $$y'(0)=e^0\cos0-e^0\sin0$$ $$y'(0)=1\times1-1\times0=1$$ 3) We know $y'(0)$, therefore the slope of the tangent line $(l)$ to $y$ at point $A(0,1)$ is also known, since it equals $y'(0)$. So the equation of the tangent line $(l)$ to $y$ at point $A(0,1)$ would be $$(l): y-1=y'(0)(x-0)$$ $$(l): y-1=1\times x$$ $$(l): y=x+1$$