Re: defining a function with D

*To*: mathgroup at smc.vnet.net*Subject*: [mg37175] Re: defining a function with D*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Tue, 15 Oct 2002 04:17:48 -0400 (EDT)*References*: <aodnti$mlg$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Jason, We have dfx[x_,t_]= D[f[x,t],x] t Cos[t x] One advantage of using = rather than := is that it differentiates once, when the definition is stored, Definition[dfx] t Cos[t x] With := we get Clear[dfx] dfx[x_,t_]:= D[f[x,t],x] Definition[dfx] dfx[x_, t_] := D[f[x, t], x] So the differentiation is done each time that the function dfx is evaluated. -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Jason Miller" <millerj at truman.edu> wrote in message news:aodnti$mlg$1 at smc.vnet.net... > Dear Listers, > > I find myself defining functions in terms of differentiation. For example, > > f[x_,t_]:=Sin[x*t] > dfx[x_,t]:=D[Sin[y,t],y]/.y->x > > This works well, but it seems to me that there should be a better way > to do this. That is, there should be a better way to define a > 'derivative' of a previous function without going through the > replacement contortions. I can't find the answer in the archive. > Can someone tell me the most straightforward way to do this? Will it > work to define a gradient vector or Jacobian matrix? A Hessian > matrix? > > Thanks in advance. > -- > Jason Miller, Ph.D. > Division of Mathematics and Computer Science > Truman State University > 100 East Normal St. > Kirksville, MO 63501 > http://vh216801.truman.edu > 660.785.7430 >