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MathGroup Archive 1998

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Re: Derivative via mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg14018] Re: Derivative via mathematica
  • From: dredd <md15987 at dredd.swipnet.se>
  • Date: Wed, 16 Sep 1998 14:11:59 -0400
  • Organization: A Customer of Tele2
  • References: <6s4rq2$f3n$3@dragonfly.wolfram.com>
  • Sender: owner-wri-mathgroup at wolfram.com

I think an expression like:

Dt[(m/(1+Exp[1/t] +b),t]/.{Dt[m,t]->p,Dt[b,t]->q}

vill do what you want, if you do just: Dt[(m/(1+Exp[1/t] +b),t] you vill
see how it works.
The /.{Dt[m,t]->p,Dt[b,t]->q}part is a replacement rule, Dt[] is total
derivative.
Peter Weijnitz
pewei at fmv.se

jpk at max.mpae.gwdg.de wrote:

> > 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)
> >
>
>




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