RE: Derivative via mathematica
- To: mathgroup@smc.vnet.net
- Subject: [mg10506] RE: [mg10486] Derivative via mathematica
- From: Ersek_Ted%PAX1A@mr.nawcad.navy.mil
- Date: Tue, 20 Jan 1998 02:22:54 -0500
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 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
|earlier.
|
I don't understand this question.
Ted Ersek