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

**References**:**Why D[x', x] isn't 0 ?***From:*"Vladimir" <vladimir347@yahoo.com>