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MathGroup Archive 2001

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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...
>



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