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Re: a problem with differentiation


Define h=2t^2 but v:=4/2(h^3+4h^2)
The SetDelayed in the definition of v should not disturb your code.
It does allows this though:

In[55]:=
Block[{h = H}, Function[{H}, D[v, h] // Evaluate]]
Out[55]:=
Function[{H}, 2(8 H + 3 H^2)]

The result is a pure Function, which eliminates the problem of h
spontaneously evaluating into 2t^2.
Orestis


"Soh Pek Hooi" <fbasohph at nus.edu.sg> wrote in message
news:9hs0sk$br6$1 at smc.vnet.net...
> 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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