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Re: Turning Derivative into Function (Newbie Question)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg115468] Re: Turning Derivative into Function (Newbie Question)
  • From: Peter Pein <petsie at dordos.net>
  • Date: Tue, 11 Jan 2011 19:23:18 -0500 (EST)
  • References: <ighgml$al6$1@smc.vnet.net>

On 11.01.2011 12:59, 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 :)
>
>
Hi stranger,

I guess, you tried g[x_]:=f'[x] while g hold still 2*x. Try Clean[g] 
before assigning a new value.

I would use
  f[x_] := x^2;
g = Derivative[1][f];
g[z]
--> 2 z

Peter


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