Evaluate higher order derivative at some point under some assumptions

I need to evaluate the $$k^{\rm th}$$ order derivative

D[w[t] w'[t], {t, k}]

which is actually equal to

Sum[Binomial[k, j] D[w[t], {t, k - j + 1}] D [w[t], {t, j}], {j, 0, k}]

at t = 0. I am given that w==0 and w'==0.

I am trying

Evaluate[D[w[t] w'[t], {t, 2}] /. t -> 0]

and then

FullSimplify[3 w'w'' + ww''', Assumptions -> w == 0 && w' == 0]

but this works only for fixed k.

Is there a way to do the evaluation at t = 0 under the asuumptions w==0 and w'==0 for arbitrary k and also take into account the value of evaluations up to k-1 order? Any help is appreciated. Thank you.

Edit

I forgot to mention that

w''[t] = f[t] - w[t] w'[t]

where f[t] is a given function. So that

D[w[t] w'[t], {t, k}] = D[w[t], {t, k + 2}] - D[f[t], {t, k}]

Thus,

w'' = f

w''' = f' - Evaluate[D[w[t]w'[t], {t, 1}]/.t->0]

etc. I need the eventual formula to take these values into account as well.

• Sum[Binomial[k, j] D[w[t], {t, k - j + 1}] D[w[t], {t, j}], {j, 2, k - 1}]? – Michael E2 Dec 14 '18 at 14:28
• Thank you! But that accounts only for w == 0 and w' == 0. What I am mainly interested in, is to account for intermediate values, like w'', w''', etc. – Asatur Khurshudyan Dec 16 '18 at 1:05
• I don't understand. Do you want to set w'' etc. equal to some given values? – Michael E2 Dec 16 '18 at 1:11
• Sorry for misleading. Your suggestion was quite obvious and evidently correct. But why the summation goes from j = 2 to j = k - 1? – Asatur Khurshudyan Dec 16 '18 at 4:26
• From the formula, one can see that those are the terms that do not contain w or w'. – Michael E2 Dec 16 '18 at 12:24