Re: Q. How to work with derivative?
- To: mathgroup at smc.vnet.net
- Subject: [mg6679] Re: [mg6615] Q. How to work with derivative?
- From: "Preferred Customer" <sherman.reed at worldnet.att.net>
- Date: Wed, 9 Apr 1997 09:15:58 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Clear[f,s]; f[x_]=x^3 s[x_]=D[Log[f[x]],x] s[3] !(x^3)(*this is output of the func def*) !(3/x)(*this is output of the deriv*) 1 (* this is the deriv at x=3 *) Although it is hard to read, I used your code and it executed just fine on 3.0 as well as on 2.2.3 Sherman Reed ---------- > From: Yaroslaw Bazaliy <yar at leland.stanford.edu> To: mathgroup at smc.vnet.net > To: mathgroup at smc.vnet.net > Subject: [mg6679] [mg6615] Q. How to work with derivative? > Date: Saturday, April 05, 1997 11:07 PM > > If any one has any experience with the following, could you please > help? > > Suppose I have a function f[x], say f[x]=x^3. Then I want to calculate > the derivative of say Log[f[x]] and evaluate it at some point, say x=3. > I try: > -------------------------------------- > Clear[f,s]; > f[x_]:=x^3; > s[x_]:=D[Log[f[x]],x]; > s[3] > General::"ivar": "\!\(3\) is not a valid variable." > Out[47]= > \!\(\[PartialD]\_3 Log[27]\) > ------------------------------------- > Try onother approach: > ------------------------------------------------ > Clear[f]; > f[x_]:=x^3; > Function[x,D[Log[f[x]],x]][3] > General::"ivar": "\!\(3\) is not a valid variable." > Out[48]= > \!\(\[PartialD]\_3 Log[27]\) > ------------------------------------------------ > Both times my "3" goes into the notation for derivative, so > it appears as if I want to differentiate with respect to "3". > > Of course I can do > ----------------------------------------------- > Clear[f]; > f[x_]:=x^3; > D[Log[f[x]],x]/.x->3 > > Out[]= > 1 > ----------------------------------------------- > and get the result, but I need a _function_ (which I can later > plot for instanse). What is the possible way to do it along the lines > of my first approach? > > Thank you for all suggestions, > Yaroslaw.