Question

Find equations of the tangent line and the normal line to the curve y = 2xe^x at the point (0,0)..

Answer 1

your derivative is: y' = 2e^x + 2xe^x. This will give you the slope of the tangent. At x=0, y' = 2. This slope along with the point (0, 0) gives y = 2x as the equation of the tangent line.

For the normal line, the slope would be -1/2. So your equation for the normal should be y = -1/2*x

Answer 2

yeah posted that too soon thanks anyway