Re: simple Problem: D[ ]

*To*: mathgroup at smc.vnet.net*Subject*: [mg27353] Re: simple Problem: D[ ]*From*: "Peter Meulbroek" <meulbroek at wag.caltech.edu>*Date*: Wed, 21 Feb 2001 03:17:59 -0500 (EST)*Organization*: California Institute of Technology, Pasadena*References*: <96t91a$pqt@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

BTW: I get different results from my calculations. In[9]:=j[x] = x^2 Out[9]=\!\(x\^2\) In[12]:= abv = D[j[x], x] Out[12]=2 x probably best to use a pattern in the definition of the function. If you want to define a function f that takes one argument, try f[x_] = x^2, or f[x_] := x^2 the first is set; it replaces the pattern f[x], where x is just about anything, with x^2. The second is setdelayed; it doesn't evaluate immediately. To understand the difference; f[x_] = x^2 g[x_] = D[f[x], x] g[x_] will have the value of 2 x The derivative is calculated, and the results are inserted into the symbol g. This works well most situations. For example, g[2 + y] will be 2*(2+y) In the setdelayed case, h[x_] := x^2 k[x_] := D[h[x], x] k[2+y] will give an error, since the x in the derivative in the definition of k evaluates to 2+y, and D[(2+y)^2, 2+y] is a syntaxt error. In summary, the symbol k is set to the UNEVALUATED expression D[]. This is useful in some contexts, not in others. "Christian Maier" <maier1 at sbox.tu-graz.ac.at> wrote in message news:96t91a$pqt at smc.vnet.net... > > I use mathmatica the first time and have a problem with a deriviation: > > in a simple problem: > > f[x]=x^2 > abl=D[f[x],x] > > then Mathematica puts out f[x]' , but not 2 x > > How does it work correctly? > Thank aou in advance! > > Christian Maier > > >