*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