## Calculus: Early Transcendentals 8th Edition

The equation of the tangent line $$(l): y=2x$$ The equation of the normal line $$(m):y=-\frac{1}{2}x$$
$$y=f(x)=2xe^x$$ 1) Find $f'(x)$ $$f'(x)=2[x'e^x+x(e^x)']$$ $$f'(x)=2(e^x+xe^x)$$ $$f'(x)=2e^x(x+1)$$ 2) The slope of the tangent line $l$ to the given curve at point $A(a,b)$ is $f'(a)$. Therefore, at $(0,0)$, the slope of tangent line $l$ is $$f'(0)=2\times e^0\times(0+1)$$ $$f'(0)=2\times1\times1=2$$ The equation of the tangent line $l$ to the given curve at $(0,0)$ has the following form: $$(l): (y-0)=f'(0)(x-0)$$ $$(l): y=2x$$ 3) The normal line is perpendicular with the tangent line. That means the product of its slopes must equal $-1$. So, the slope of the normal line $m$ at $(0,0)$ is $\frac{-1}{2}$ The equation of the normal line at $(0,0)$ has the following form: $$(m):(y-0)=-\frac{1}{2}(x-0)$$ $$(m):y=-\frac{1}{2}x$$