Differentiate k*t*e^-t twice use chain rule or product rule

Hi,
I'm stuck trying to differentiate, am I supposed to use the product rule: uv'+vu' or the chain rule, I need to differentiate once and take that answer and differentiate again. any help would be very useful! Thanks

Let y = kte^-t.............where k is a constant

Then, y =k(uv)......where u = t and v = e^-t

so, u' = 1 and v' = -e^-t

=> y' = (t)(-e^-t) + (1)(e^-t) = -te^-t + e^-t = k(-te^-t + e^-t)

Differentiating again gives:

y'' = (-t)(-e^-t) + (-1)(e^-t) - e^-t = te^-t - 2e^-t => ke^-t(t - 2)

:)>

lets assume that we will differentiate k*t*e^-t respect to t..

f(t) = kte^(-t)
f'(t) = k[e^(-t) d/dt 1 + t d/dt -e^(-t)]
=k[e^(-t) - te^(-t)]
= ke^(-t) - kte^(-t)

now solving the 2nd derivative..

f''(t) = -ke^(-t) - k[e^(-t) d/dt 1 + t d/dt -e^(-t)]
=-ke^(-t) -ke^(-t) + kte^(-t)
=-2ke^(-t) + kte^(-t)
=ke^(-t)[t - 2] answer//

k*t*e^-t

Differentiating once,
k*(t*e^-t*-1 + e^-t)

Differentiating a second time,
k*[-(t*e^-t*-1 + e^-t) + e^-t]

= k*[t*e^-t - e^-t + e^-t]
= k*t*e^-t

The same thing.

holy freaking nerd talk!!! lol you guys are brillaint, or at least it looks like it :)