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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.
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University of Massachusetts                413 545-2859 (W)
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