RE: Derivative of multiargument function
- To: mathgroup at smc.vnet.net
- Subject: [mg29700] RE: [mg29678] Derivative of multiargument function
- From: "David Park" <djmp at earthlink.net>
- Date: Wed, 4 Jul 2001 03:08:25 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Apparently, Mathematica did not evaluate the derivative because the second slot in your function was not defined as a variable slot. Using a slightly more interesting function... Clear[g] g[z_, i_] := z^i Derivative[1, 0][g][z, 0] 0 Derivative[1, 0][g][z, i] i*z^(-1 + i) But when an expression for a function contains symbols that are clearly separated between "parameters" and "variables", I usually prefer to define the function this way... Clear[g] g[i_][z_] := z^i Derivative[1][g[0]][z] 0 Derivative[1][g[i]][z] i*z^(-1 + i) Or even more directly g[0]'[z] 0 g[i]'[z] i*z^(-1 + i) David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ > From: guido [mailto:carlet at sissa.it] To: mathgroup at smc.vnet.net > > If I have the assignement: > g[z_] = 0 > then any time I have: > Derivative[1][g][z] > I obtain zero and this is ok. > But if I have a sequence of functions with j=1,2,... > and I give the assignment: > g[z_,1] = 0 > then Mathematica doesn't seem to give zero when it finds: > Derivative[1,0][g][z,1] > as before... > does anybody know how to let Mathematica compute the derivative even > in this case ? > Thank you very much... >