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]==0 and w'[0]==0.

I am trying

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

and then

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

but this works only for fixed k.

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


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}]


w''[0] = f[0]

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

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

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

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