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RE: Derivative via mathematica    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.
|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) |
|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? |

I think this does what you want.

 expr=Dt[  m/(1+Exp[1/t] +b),  t ]

Dt[m, t]/(1 + b + E^(1/t)) -
  (m*(Dt[b, t] - E^(1/t)/t^2))/(1 + b + E^(1/t))^2


p/(1 + b + E^(1/t)) - (m*(q - E^(1/t)/t^2))/
   (1 + b + E^(1/t))^2


"Dt[f, x] gives the total derivative of f with respect to x. Dt[f] gives
the  \
total differential of f. Dt[f, {x, n}] gives the nth total derivative of
f \ with respect to x. Dt[f, x1, x2, ... ] gives a mixed total

(*  You also asked about the folllowing:   *) |
|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? |
|Basically, it is a chain derivative question, I just want it to stop

I don't understand this question.

     Ted Ersek

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