Answer
The equation of the tangent line $$(l): y=2x$$
The equation of the normal line $$(m):y=-\frac{1}{2}x$$
Work Step by Step
$$y=f(x)=2xe^x$$
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: $$(m):(y-0)=-\frac{1}{2}(x-0)$$
$$(m):y=-\frac{1}{2}x$$
