# Finding the equation of a tangent line at a given point

Doing my final review for the semester and just want to make sure I have all my bases covered. (I'll more than likely have another 4-5 question to come today, but I'll start here).

Looking for someone to confirm whether this is the correct answer or not. If not, can you explain where I've gone wrong.

Find the equation of the tangent line to $y = 2xe^{-x}$ at the point (0,0)

1 (find derivative) :

$$y' = 2(e^{-x}) + 2x(-e^{-x})$$

2 (plug in x) :

$$2(e^{-0}) + 2(0)*(-e^{-0}) = 2(1) + 2(0)*(-1)$$

3 (plug that in to point slope formula) :

$$y - 0 = 2(x + 0) --> y = 2x$$