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RE: a problem with differentiation


Hi Carlos, thanks. 

I tried your recommended method and it worked fine. But when I tried to put
it in the following format, it said 2t^2 is a not a valid variable.

h[t_] := 2t^2
v[h_] := 4/2(h^3 + 4h^2)
Function[{h,v}, D[v[h], h]][2t^2, 4/2(h^3+4h^2)]

We have a program which requires the Function to be specified so that any
expressions for h and v can be passed in. How can we still "hold" on to "h"
without h being substituted in the Function?

Pek

-----Original Message-----
From: Carlos Collier
To: mathgroup at smc.vnet.net
Subject: [mg29717] Re: [mg29676] a problem with differentiation

On Tue, 3 Jul 2001, Soh Pek Hooi wrote:
Hello

you could try the following:
h[t_] := 2t^2
v[h_] := 4/2(h^3 + 4h^2)

D[v[h], h]

   2*(8*h + 3*h^2)

saludos
carlos


> Hi,
>
> If I have
> h= 2t^2
> v = 4/2(h^3+4h^2)
>
> How do I evaluate D[v,h] in terms of h without having h to be
substituted in
> the following manner?
>
> Function[{h,v}, D[v, h]][2t^2, 4/2(h^3+4h^2)]
>
> Will appreciate your comments.
>
> Pek
>
>


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