Partial differentiation help! (10 pts)

Hi there, i've a question which i'm struggling with and would appreciate any help given (with working):
I need to partially differentiate the below function TWICE with respect to both x and then t.
So what i'm looking for is : dy/dx, d2y/dx2, dy/dt and d2y/dt2.

y(x,t) = 2 x e^-x-3t + 6 t e^-x-3t

Many thanks to anybody who helps, i really appreciate it :)

To compute, the partial derivatives of y(x, t) = [(2x)e^(-x - 3t) + (6t)e^(-x - 3t)], one should rewrite it as:

y(x, t) = {2x + 6t}e^(-x)e^(-3t)

For ∂[y(x, t)]/∂x you view e^(-3t) as a constant and use the product:

e^(-3t)∂[g(x, t)h(x)]/∂x = e^(-3t)[{∂g(x, t)/∂x}h(x) + g(x){dh(x)/dx}]

let g(x, t) = 2x + 6t, then ∂g(x, t)/∂x = 2
let h(x) = e^(-x), then dh(x)/dx = -e^(-x)

e^(-3t)∂[g(x, t)h(x)]/∂x = e^(-3t)[{2}e^(-x) + (2x + 6t){-e^(-x)}]

∂y(x, t)/∂x = (2 - 2x - 6t)e^(-x)e^(-3t)

You use the product rule, again, to compute the second derivative:

let g(x, t) = 2 - 2x - 6t, ∂g(x,t)/∂x = -2
let h(x) = e^(-x) then h'(x) = -e^(-x)

∂²y(x, t)/∂x² = -2e^(-x)e^(-3t) - (2 - 2x - 6t)e^(-x)e^(-3t)

∂²y(x, t)/∂x² = (2x + 6t - 4)e^(-x)e^(-3t)

You do the same for partials with respect to t:

∂y(x, t)/∂t = 6e^(-x)e^(-3t) + -3(2x + 6t)e^(-x)e^(-3t)

∂y(x, t)/∂t = 6e^(-x)e^(-3t) + (- 6x - 18t)e^(-x)e^(-3t)

∂y(x, t)/∂t = (6 - 6x - 18t)e^(-x)e^(-3t)

∂²y(x, t)/∂t² = -18e^(-x)e^(-3t) + -3(6 - 6x - 18t)e^(-x)e^(-3t)

∂²y(x, t)/∂t² = (54t + 18x - 36)e^(-x)e^(-3t)

If you like, you can change e^(-x)e^(-3t) back to e^(-x - 3t) in all of the answers.