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How can I directly evaluate the following derivative? $$ \frac{\partial ^n}{\partial x^n}\bigg[\sum_{k=1}^{m} (m-k)! (ax+b)^k\bigg] = \sum_{k=1}^{m} (m-k)! \bigg[\frac{k!}{(k-n)!}a^n (ax+b)^{k-n}\bigg] $$

Mathematica code is

D[Sum[(m - k)! (ax + b)^k, {k, 1, m}], {x, n}] // FullSimplify

or even this will do:

D[(ax + b)^k, {x, n}] // FullSimplify

EDIT 1:

nthDeriv[f_,x_,n_]:=n!*SeriesCoefficient[f[x],{x,x,n}]
f2[x_] := (a + b x )^k
nthDeriv[f2, x, m]

I obtain a weird output, however, for

f1[x_] := (1 +  x )^n

the above method works fine.

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    $\begingroup$ Have a look at SeriesCoefficient. $\endgroup$ Jan 20 '19 at 9:00
  • $\begingroup$ @b.gatessucks SeriesCoefficient seems to be yielding weird output. $\endgroup$ Jan 20 '19 at 9:44
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D[(a x + b)^k, {x, n}]

is straightforward:

$$a^n k^{(n)} (a x+b)^{k-n}$$

Note: don't use ax when you should use a x.

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    $\begingroup$ Thanks for pointing out. But mathematica 10.04 does not yield the desired output. Which version are you using? $\endgroup$ Jan 21 '19 at 5:03
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    $\begingroup$ I'm using 11.3.0.0 on a Mac. $\endgroup$ Jan 21 '19 at 5:11

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