# 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
____________________________________________________________________

```

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