Re: Turning Derivative into Function (Newbie Question)

*To*: mathgroup at smc.vnet.net*Subject*: [mg115473] Re: Turning Derivative into Function (Newbie Question)*From*: Tomas Garza <tgarza10 at msn.com>*Date*: Tue, 11 Jan 2011 19:24:18 -0500 (EST)

If g is to work as a function it should have an argument, which it doesn't in your first definition.When redefining it, you should first Clear the previous definition. Then you get In[1]:= Clear[g]In[2]:= g[x_]:=f'[x] In[3]:= g[2]Out[3]= 4 as expected. -Tomas > Date: Tue, 11 Jan 2011 06:58:59 -0500 > From: forpeopleidontknow at gmail.com > Subject: [mg115451] Turning Derivative into Function (Newbie Question) > To: mathgroup at smc.vnet.net > > 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 :) >