Calculus question! 10 points

Hello users,
Could anyone please tell me step by step on how to approach and solve a problem like this.
y=x^2*e^2x ; I need to find the Tangent on this graph at x = 1.
I know that I have to differentiate y then put in 1 and use y-y1=m(x-x1), BUT could someone please take me thorough those steps. All help is appreciated.
Thanks in advance.

y = x^2 * e^(2x)
dy/dx = x^2 * 2 * e^(2x) + 2x * e^(2x)
dy/dx = 2x * e^(2x) * (x + 1)
x = 1
dy/dx = 2 * 1 * e^(2 * 1) * (1 + 1)
dy/dx = 2 * 2 * e^2
dy/dx = 4e^2

y = x^2 * e^(2x)
y = 1 * e^(2 * 1)
y = e^2

y = mx + b
e^2 = 4e^2 * 1 + b
-3e^2 = b

y = 4e^2 * x - 3e^2
y = (e^2) * (4x - 3)

hmm i dont' exactly know how to read that equation. is it y=x^(2*e^2x) or y=(x^2)(e^2x)
either way, you find the derivative of y, and plug in x=1. this will get you the slope at x=1.
since you know the slope, you can figure out the value of y at x=1. then you can find the y-intercept. then you can find the tangent line.