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One can expand the hypergeometric function as a series of the last argument and take the derivative series[Derivative[n__][f_][args__], k_] := Module[{vars = {args} /. Except[_List | List] :> Unique[]}, FullSimplify[# (Last@vars)^k /. Thread[Flatten@vars -> Flatten@{args}], Assumptions -> {k ∈ Integers, k >= 0}] &@ ...


This isn't exactly an answer but perhaps it's a step in the right direction. Actually with the subsequent edits I think it is an answer. You'll need to do the calculations yourself and check that I didn't screw anything up but I think this works and gives you a closed form expression. f[x_] = FunctionExpand[HypergeometricPFQRegularized[{1, 1, 1, 1}, {2, 2, ...


Here is the "code" of your question with unspecified blanks filled in and errors fixed: F[u_] := Log[u] - Cos[u^2] Exp[-u]; Umax[x_, y_] := x + y; W[x_?NumericQ, y_?NumericQ] := NIntegrate[F[u], {u, 0, Umax[x, y]}]; g[x_?NumericQ, y_?NumericQ] := x^2/y^2; Alpha[x_?NumericQ] := Tanh[x]; myFunction[x_?NumericQ] := NIntegrate[g[x, y]*W[x, y], {y, ...


Unprotect[Dot]; SetAttributes[Dot, Orderless] Protect[Dot] then try to evaluate this again D[V[t]/Sqrt[Dot[V[t], V[t]]], t] of course it would be better to define new function based on dot and set it attribute Orderless

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