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RE: Derivative via mathematica
ccai1@ohiou.edu 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.
In[1]:=
expr=Dt[ m/(1+Exp[1/t] +b), t ]
Out[1]=
Dt[m, t]/(1 + b + E^(1/t)) 
(m*(Dt[b, t]  E^(1/t)/t^2))/(1 + b + E^(1/t))^2
In[2]:=
expr/.{Dt[m,t]>p,Dt[b,t]>q}
Out[2]=
p/(1 + b + E^(1/t))  (m*(q  E^(1/t)/t^2))/
(1 + b + E^(1/t))^2
In[3]:=
?Dt
"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
derivative."
(* You also asked about the folllowing: *) 
A related question, I tried to use nondefined 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
earlier.

I don't understand this question.
Ted Ersek
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