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

```

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