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MathGroup Archive 2001

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RE: The value of a partial derivative

  • To: mathgroup at smc.vnet.net
  • Subject: [mg26737] RE: [mg26717] The value of a partial derivative
  • From: "David Park" <djmp at earthlink.net>
  • Date: Fri, 19 Jan 2001 02:14:13 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Anastasius,

Here is one method, using substitution rules:

expr = a*x + b*x^2;

D[expr, x]
% /. {a -> 1, b -> 2, x -> 3}
a + 2 b x
13

Here is another method, using a function definition for the original
expression. By using separate square brackets for the parameters, (a, b),
and the variable, x, you can use the standard Mathematica derivative
notation.

f[a_, b_][x_] := a*x + b*x^2
f[1, 2]'[3]
13

David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/


> From: Anastasius Moumtzoglou [mailto:anas1 at hol.gr]
To: mathgroup at smc.vnet.net
> Hi,
>
> I have derived the partial derivative of an expression and I would like
> to give values to the parameters in order to get a value for the partial
> derivative. Is there a way I could do it using Mathematica 3.0?
>
> ----------------------------------------------------------------------
> Anastasius Moumtzoglou
> 1) Educational & Tecnological Institutions,
> Department of Health Services Management
> 2) National School of Public Health,
> Department of Health Services Management
> 3) "P. & A. Kyriakou" Children's Hospital
> Personal Page: http://www.geocities.com/moumtzoglou/
> Email:anas1 at hol.gr
>
>



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