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Re: Why D[x', x] isn't 0 ?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg66116] Re: [mg66094] Why D[x', x] isn't 0 ?
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Sun, 30 Apr 2006 04:21:05 -0400 (EDT)
*References*: <200604290740.DAA23451@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On 29 Apr 2006, at 16:40, Vladimir wrote:
> Hello!
>
> D[x'', x'] => 0 (* okay *)
>
> D[x', x] => x'' (* unexpected despite being similar to above *)
>
> D[f[x'], x] => x'' f'[x'] (* unexpected as well *)
>
> Why the last 2 lines don't produce zeroes and how to make them
> do so (for any function f) because it's quite common in mechanics
> where velocity is treated as independent of position, etc.
>
> --
> Vladimir
>
Consider the case D[x', x]. Mathematica does it as follows. The
FullForm of the expression is:
D[Derivative[1][x], x]
So Mathematica sees it as having the form f[x], where f is the
function Derivative[1]. Next, Derivative[1] is differentiated as a
function to obtain Derivative[2]. Finally Derivative[2][x] is
returned and re-written as x''.
Now the case D[x'', x']
The FullForm is D[Derivative[2][x], Derivative[1][x]]. Now the
expression that is being differentiated is not seen as an explicit
function of the expression Derivative[1][x] with respect to which
the differentiation is being performed, so the answer 0 is returned.
The other case should now be obvious.
As for your other question: the easiest way is the most obvious one:
In[1]:=
x'=v;
In[2]:=
D[f[x'], x]
Out[2]=
0
In[3]:=
D[x',x]
Out[3]=
0
There are other, more elaborate ways, but I can't imagine that they
would be preferable to the above.
Andrzej Kozlowski
Tokyo, Japan
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