# Expand double derivative when substituted?

Posted 11 months ago
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 I am trying to evaluate: Expand[D[x[t], {t, 2}] + (x[t])^3 /. x[t] -> A sin[[Omega]t]] in Mathematica. However, I get the following output: A^3 sin[[Omega]t]^3 + (x^[Prime][Prime])[t] whereas I am expecting and output where x''[t] is also differentiatedCould somebody help me figure out what steps to follow to solve this issue?
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Posted 11 months ago
 Try:  ClearAll["Global*"]; Remove["Global*"]; (x''[t] + (x[t])^3) /. x[t] -> A Sin[?[t]] /. x''[t] -> D[A Sin[?[t]], {t, 2}] (*A^3 Sin[?[t]]^3 + A (-Sin[?[t]] Derivative[1][?][t]^2 + Cos[?[t]] (?^??)[t])*) Or: ClearAll["Global*"]; Remove["Global*"]; x[t_] := A Sin[?[t]]; D[x[t], {t, 2}] + (x[t])^3 (*A^3 Sin[?[t]]^3 + A (-Sin[?[t]] Derivative[1][?][t]^2 + Cos[?[t]] (?^??)[t])*) 
 You could also try the anonymous function (i.e., #&) approach where you replace x[t] with # like so (D[#, t, t] + #^3) &[A Sin[Omega t]] (* -(A*Omega^2*Sin[Omega*t]) + A^3*Sin[Omega*t]^3 *) 
 You have to replace the symbol x with a function: x''[t] + x[t]^3 /. x -> Function[t, A*Sin[\[Omega]*t]] The pattern x[t_] does not match anything in x''[t].