Re: Derivative via mathematica
- To: mathgroup@smc.vnet.net
- Subject: [mg10566] Re: Derivative via mathematica
- From: Paul Abbott <paul@physics.uwa.edu.au>
- Date: Tue, 20 Jan 1998 16:54:08 -0500
- Organization: University of Western Australia
- References: <69ncl9$8d5@smc.vnet.net>
cai wrote:
> I just used mathematica for a couple of days. I am trying to compute
> the derivative under mathematica. Because the function is complicated,
> I like to break it down.
>
> f[t_] = (m/(1+Exp[1/t] +b)
>
> Here m and b are functions of t.
Then you probably should enter this as
In[1]:= f[t_] = m[t]/(1 + Exp[1/t] + b[t])
Out[1]=
m[t]
-----------------
1
b[t] + Exp[-] + 1
t
> What I want is if I define the values of m' and b', rewrite the f
>
> m' = p
> b' = q // well, I dont know how to define, this is the idea
>
> f[t_, m[t], b[t]] = (m/(1+Exp[1/t] +b)
>
> then use the command D[f[t,m[t],b[t]],t] hopeful get a equation which is
> the function of t, p and q. How to do that?
Is this what you want (using pattern-matching and replacement rules)?
In[2]:=f'[t]
Out[2]=
1
Exp[-]
t
m[t] (b'[t] - ------)
2
m'[t] t ----------------- -
---------------------
1 1 2 b[t] + Exp[-] + 1 (b[t] +
Exp[-] + 1)
t t
In[3]:= %/.f_[t_]:>f/.{m'->p,b'->q}
Out[3]=
1
Exp[-]
t
m (q - ------)
2
p t
-------------- - -----------------
1 1 2
b + Exp[-] + 1 (b + Exp[-] + 1)
t t
Cheers,
Paul
____________________________________________________________________
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia Nedlands WA 6907
mailto:paul@physics.uwa.edu.au AUSTRALIA
http://www.pd.uwa.edu.au/~paul
God IS a weakly left-handed dice player
____________________________________________________________________