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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