Re: Turning Derivative into Function (Newbie Question)

*To*: mathgroup at smc.vnet.net*Subject*: [mg115457] Re: Turning Derivative into Function (Newbie Question)*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Tue, 11 Jan 2011 19:21:10 -0500 (EST)

Because g is NOT a "function" when you set: g := f'[x] (No more than f would be had you started with f := x^2.) Whenever you now call g, you get the value of f[x], namely, 2x. If you really want to define g that way, then you can use a rule and replacement: g /. x-> 2 But it's probably better to do things like this: Clear[g] g[x_] = f'[x] (* note Set, =, not SetDelayed, := *) g[2] 4 Or, if you prefer a more functional approach: Clear[g] g = f' g[2] 4 On 1/11/2011 6:58 AM, Just A Stranger wrote: > When I apply a function that outputs another function, how come I can not > pass arguments to the new function? Am I missing some principle about how > Mathematica functions work? Could someone please point me to the relevant > documentation? > > > A simple example to illustrate my confusion: > > In[1]: f[x_] := x^2 > In[2]: g := f'[x] > > So now I have: > > In[3]: g > out[3]: 2x > > as expected > > So how come I get the following when trying to pass an argument to g: > > In[4]: g[2] > Out[4]: 2x[2] > > Instead of the output I want: > > *Ou[4]: 4 > > > I tried > > In: g[x_] := f'[x] > > But it seems to think I'm trying to assign a function to Times. > "SetDelayed::write: Tag Times in (2 x)[y_] is Protected.>>" > > Thank you very much for any help :) > > -- Murray Eisenberg murray at math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305