Talk:Total derivative

message: At present, the article says: "in multiplication, the total relative error is the sum of the relative errors of the parameters."
but I'd think this is not completely correct... for this to be true in general, don't we need to assume the errors of the parameters to be independent, and indeed to further qualify that this is correct true to first-order for small relative errors (i.e., it is a linearization in the size of the fractional error)?  But I don't dare to correct the article myself since I'm not confident in my statistical abilities...  --67.174.192.178 07:54, 13 September 2005 (UTC)
replies: text: You are right that this depends on the underlying assumptions of the error model (it's also a worst-case analysis). Ideally, there should be an article somewhere explaining this that we could link to, but I cannot find one. It might be a bit too much to explain the error model (which is by the way the usual one in measurements, I think) in this article. Do you have any suggestions on the best way to tackle this? -- Jitse Niesen (talk) 13:38, 13 September 2005 (UTC)
thread_title:  Relative error 

message: I've made the existing material on this page mathematically correct (I hope) and added some material. I don't think the page is yet complete, but this is all I want to do for the time being. Geometry guy 18:44, 25 February 2007 (UTC)
replies: text: (Well, I added a few minor points. I noticed that some authors/editors distinguish between (total) derivative and (total) differential. Is this widespread? I mentioned both terminologies anyway. Geometry guy 23:48, 20 March 2007 (UTC)

text: The current version is quite clean until the Application of the total derivative to error estimation section, where the formulae need to be correctly formatted into tex. 163.1.14.45 07:19, 25 July 2007 (UTC)

text: The first two bullet points appear to be the same. Moreover, the introduction is written like a (very brief and unclear) first section, or a couple of sections. If the differences are that pronounced, shouldn't this have a disambiguation page? Also, the synonyms should be removed or put under a "Other meanings" type section. Do we know who wrote the original page?  — Preceding unsigned comment added by 86.163.9.39 (talk) 19:41, 27 September 2013 (UTC)
thread_title:  Clean-up 

message: Can somebody use simple sentences to define total derivative? The definition is very confusing with too many punctuations.
128.233.111.111 (talk)  —Preceding comment was added at 23:53, 28 November 2007 (UTC)
replies: text:  I've modified everything (in the first definition) to be in terms of t. There were still some z's in the equation that were not mentioned anywhere else. Note that the differential operator is *not* using total derivatives WITHIN the operator, while the explicit definition is. Maybe someone should clear that up. --TedPavlic | talk 13:38, 20 December 2007 (UTC)
thread_title: Confusing Definition

message: What do you think of my section User:Saippuakauppias/Fréchet (in fact, I samw this article much after. However, I think, in my version it's quite clear and more explicit, then here. What do you think of? - Please post your comments to the articles Talk site. Thanks for your comments! --Saippuakauppias ⇄ 19:22, 28 April 2008 (UTC)
thread_title:  Same definition, in an other way - more clear? 

message: I would like to see a diagram and some discussion of the intuitive meaning of the total derivative. It seems like it comes up when I have a function of several variables with constraints among those variables. (I can't see what it means without constraints among the arguments.) If I have f(x,y) and constrain it with x=g(y), then f(x,y) is a surface over a 2D domain and the constrained problem is a line on that surface. The partial derivative, \partial f(x,y)/\partial y, is just the slope of f in the y direction. The total derivative of the constrained problem is:
: \frac{\mathrm{d}f(x,y)}{\mathrm{d} y} = \frac{\partial f(x,y)}{y}+\frac{\partial f(x,y)}{\partial x}\frac{\mathrm{d} x}{\mathrm{d} y}
which is the slope along the constrained line (right?). What about the case when there is no constraint: when you just have a surface over an n-dimensional domain? Does the total derivative have meaning?
A diagram of this setup with f(x,y) would be helpful to show this distinction. —Ben FrantzDale (talk) 14:57, 12 December 2008 (UTC)
thread_title:  Intuition, diagram? 

message: What this page describes is the normal Liebniz derivative. The author(s) appear to have missed the point of the chain rule! I've put in a page delete request. 202.173.164.129 (talk) 11:34, 1 April 2012 (UTC)
thread_title:  Redundant Page 

message: I think the importance of total derivatives or divergences as they are called in the Calculus of Variations is not fully demonstrated here.  Certainly for any parameter invariant Lagrangian (or Lagrangian density) where each of the coordinate parameters are on an equal footing, the following holds true:
:
\int_{x_1=-\infty}^{x_1=\infty} \int_{x_2=-\infty}^{x_2 =\infty} \ldots 
\int_{x_i=-\infty}^{x_i=\infty} \ldots \int_{x_j=-\infty}^{x_j=\infty} \ldots \int_{x_n=-\infty}^{x_n=\infty}
\frac{\partial F(x_1, x_2, \ldots x_i, \ldots x_j,\ldots x_n)}{\partial x_j} ~ d x_1 d x_2, \ldots d x_i  \ldots d x_j \ldots d x_n 
:
~=~
\int_{x_1=-\infty}^{x_1=\infty} \int_{x_2=-\infty}^{x_2 =\infty} \ldots 
\int_{x_i=-\infty}^{x_i=\infty} \dots
 \int_{x_n=-\infty}^{x_n=\infty}
\lim_{x_j \rightarrow \infty} (F(x_1, x_2, \ldots x_i, \ldots x_j, \ldots x_n ) - F(x_1, x_2, \ldots x_i, \ldots -x_j, \ldots x_n ) ) ~ d x_1 d x_2, \ldots d x_i  \ldots d x_n 
:
 ~=~ 0  ~.
This is often used and exploited in quantum field theory and General Relativity.  I think an explanation along these lines needs to be included.  Integration over an infinite range usually demands that "surface" terms go to zero.  The order of integration does not matter in the most general cases of applicability to real problems. TonyMath (talk) 03:10, 9 April 2012 (UTC)
replies: text: No response so far.  If nobody objects, can I not simply add this material to the article itself?TonyMath (talk) 02:10, 23 April 2012 (UTC)
replies: text: ok I put it in and added a reference.  Let's see if anyone objects :-)  TonyMath (talk) 01:15, 24 April 2012 (UTC)
replies: text:  Objection! I do not think this is helpful at all. It is unclear what you want to tell with this. To me almost no word you are writing makes sense here. If you can clarify this, we can certainly discuss weather this is helpful in the article. For now, I would remove this part. Regards. Falktan (talk) 17:18, 19 October 2012 (UTC)
thread_title: Simpler more general demonstration?

message: The intro claims, without a source, that "total derivative" can mean the same thing as "total differential". That's just not right--a total derivative is, roughly speaking, the ratio of differentials. So for example if 
:x = f(y(t), t)
then the total differential of f is 
:fyytdt + ft dt, 
while dividing this through by dt gives the total derivative dx/dt.
This has been mentioned a couple of times in years past on this talk page. Since no one has disagreed and no one has put in a reference refuting this point, I'm taking out the false assertion. Loraof (talk) 16:46, 12 May 2015 (UTC)
replies: text: I just put at the start . It would be nice if someone added a section at the end explaining the difference, as is done briefly by Loraof above, but fleshed out a bit more. --editeur24 (talk) 04:26, 15 December 2020 (UTC)
thread_title:  A total derivative is not a total differential 

message: d(f(z(x,y)))/dz = ? 
Just granpa (talk) 20:44, 21 November 2017 (UTC)
thread_title:  how to find df/dz if z is a function of x and y? 

message: Hi, I came here to review total derivative idea after watching geodynamics and in it they wrote total derivative in a 'strange' way. I thought someone more knowledge could explain why continuity equation is expresses in the form df/dt = partial-wrt-time(f)+∆.(fv) , with the dot product. After applying Leibnitz product we have an extra term f(∆.v) hence this is not like total derivative in the article. My suspicion is that the first definition of tot derivative is the most general and explains this continuity equation . the familiar total derivative without the extra term is due to f domain being flat(R^m) . proving this is a challenge , I am not mathematician , I am just user. Can someone please include this subtlety as I think it is useful for applications petite (talk) 23:03, 29 January 2022 (UTC)
thread_title:  Conservation of mass(continuity equations) 

message: Navier stoke equation can be thought of as total derivative of momentum density equaling body forces - which is analogous to newtons law- provided the definition of total derivative is extended such that after partial derivative wrt time , we have diffusive terms minus convective time:
Total derivative = partial-wrt-time + diffusion - convectiin.
d(pv)/dt= partial-wrt-t(f)+v.∆(pv) -∆.S=f ,, S is stress, f is body forces petite (talk) 23:37, 29 January 2022 (UTC)
thread_title:  Application to Navier-stoke equation 

message: The page gives the definition as
:f(a + h) = f(a) + df_a(h) + \varepsilon(h),
where we have
:\varepsilon(h) = o(\lVert h\rVert),
but this would appear to not make sense as we must have \varepsilon : U \to \R^m. Should it perhaps be \lVert\varepsilon(h)\rVert = o(\lVert h\rVert)? I'm rusty on my analysis and so wanted another set of eyes before making a change.
Chessturo (talk) 22:38, 31 January 2024 (UTC)
thread_title:  Small error in definition?