Re: Turning Derivative into Function (Newbie Question)

*To*: mathgroup at smc.vnet.net*Subject*: [mg115470] Re: Turning Derivative into Function (Newbie Question)*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Tue, 11 Jan 2011 19:23:42 -0500 (EST)

You didn't make g a function, so it doesn't act like a function. What you apparently meant is f[x_]:=x^2 g[x_]:=f'[x] or you could replace the second statement with g = f' 2 #1 & (The last line is output of the line before it.) Bobby On Tue, 11 Jan 2011 05:58:59 -0600, Just A Stranger <forpeopleidontknow at gmail.com> 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 :) > > -- DrMajorBob at yahoo.com