Re: Derivative via mathematica
- To: mathgroup@smc.vnet.net
- Subject: [mg10573] Re: [mg10486] Derivative via mathematica
- From: jpk@max.mpae.gwdg.de
- Date: Tue, 20 Jan 1998 16:54:14 -0500
> Hi,
>
> 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.
> If I directly use command D after insert m and b terms, a very
> complicated equaion is gerenated, which I do not want.
>
> 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)
>
Hi,
The definitions:
f[t_,m_[t_],b_[t_]]:=(m[t]/(1+Exp[1/t] +b[t]))
Derivative[1][mu][t_]:=phi[t]
Derivative[1][beta][t_]:=xi[t]
will do what You want
D[f[t,mu[t],beta[t]],t] //InputForm
phi[t]/
(1 + E^t^(-1) + beta[t]) -
(mu[t]*(-(E^t^(-1)/t^2) +
xi[t]))/
(1 + E^t^(-1) + beta[t])^2
> 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?
>
> A related question, I tried to use non-defined function In[19]:= m[t_]
> Out[19]= m[t_]
> In[20]:= D[m[t],t]
> Out[20]= m'[t]
> and expected D[f[t,m[t],b[t]],t] contains m'[t]. Is it possible?
>
If You drop the definitions for Derivative[1][mu][_] and
Derivative[1][beta][_] You get the mu'[t] amd beta'[t] output.
Hope that helps
Jens
> Basically, it is a chain derivative question, I just want it to stop
> earlier.
>
> Could somebody help?
> Thanks.
>
> ccai1@ohiou.edu
>
>
>