D and sinh'; s/c vs. s*c^(-1).
- To: mathgroup at yoda.ncsa.uiuc.edu
- Subject: D and sinh'; s/c vs. s*c^(-1).
- From: fateman at peoplesparc.Berkeley.EDU (Richard Fateman)
- Date: Sun, 9 Dec 90 17:20:49 PST
I suspect that cosh'[....] := .... defines the derivative with respect to its argument, in this case, 2x, as 2*sinh[2*x]. But to take the derivative with respect to x, you get 2*sinh[2*x] * D[2*x,x] or 4. As for c^n_/s^n_ -> t^(-n), I think your pattern has a division in it, but your expression has a multiplication. c^n*s^(-n). I think you need a rule for c^n_*s^m_ -> somefunctionof(n,m). Good luck. Richard Fateman