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Thanks to all the writers of this page! Saved my behind (-:

Thanks again,Sam Krupa

Agreed. I've never known what a partial derivate is so I decieded to check it up. Clearly and well written.

---—Precedingunsigned comment added by88.114.90.71 (talk)09:43, 18 March 2009 (UTC)[reply]

Yeah, this is really an impressive piece of work. The introduction is really accessible, contrary to loads of the maths articles on wikipedia.My 2 Cents' Worth (talk)10:42, 3 June 2010 (UTC)[reply]

i came to talk to say same things- best intro  in any wilpedia item on math— Precedingunsigned comment added by174.24.33.155 (talk)19:52, 25 December 2015 (UTC)[reply]

Since there is only one variable in the formula for the area of a circle, as the article mentions, isn't the example using it a bad one?(Unsigned comment by 68.78.139.53 on17 May2006)

Although it's been many, many years since I've done any maths, I would agree - this example does seem pointless. --A bit iffy09:54, 11 June 2006 (UTC)[reply]

I agree. I've removed it. --Spoon!02:22, 11 September 2006 (UTC)[reply]

The third and the fourth paragraph, that is the discussion about the total derivative and the Jacobian are way out of place. Those things are inserted in the middle of the discussion about the partial derivatives, and that is inappropriate.

All that stuff needs to go in the last section, where the gradient is discussed. Other opinions?Oleg Alexandrov 02:27, 23 Mar 2005 (UTC)

Yes, I agree. It appears those paragraphs have already been deleted.— Precedingunsigned comment added byNickalh (talk •contribs)07:18, 24 January 2024 (UTC)[reply]

I'd like to see a section addressing the semantic difference between ∂ andd. For example, ify is only a function ofx, then what is the difference between∂y/∂x anddy/dx. Also, in sloppy engineering notation I've seendy appear alone; I think I've seen∂y alone as well. This may be rigerous use ofInfinitesimals or it may be engineering shorthand; I'm not sure. —BenFrantzDale02:00, 5 December 2005 (UTC)[reply]

I would agree with such a section, but preferably at the bottom of the article; otherwise I would think that it may confuse more than illuminate. Some connections with theLeibniz notation may be made.Oleg Alexandrov (talk)02:59, 5 December 2005 (UTC)[reply]

Now I see, the page I needed to look at wastotal derivative. —BenFrantzDale17:40, 15 December 2005 (UTC)[reply]

What actually is the difference between dy/dx and δy/δx? Is it just preference?My 2 Cents' Worth (talk)10:44, 3 June 2010 (UTC)[reply]

δy/δx can denote afunctional derivative. —winggundam03:17, 23 October 2019 (UTC)[reply]

Here is my site with partial derivative example problems. Someone please put this link in the external links section if you think it's helpful and relevant.Tbsmith

http://www.exampleproblems.com/wiki/index.php/PDE:Integration_and_Seperation_of_variables

I am a bit perplexed as to why the formal definition is at the end and the example is at the beginning, I think some reordering is in order here!Retardo22:13, 5 May 2006 (UTC)[reply]

Incomputational mechanics, it is common to denote the surface of a volume asS=∂V. My sense is that this has some underlying meaning indifferential geometry. Is it just a notational convenience or is there some rigerous extension of partial differentiation onto volumes? Also, how does this relate to thedV you find in a volume integral? (TheredV basically means "one tiny point inV" whereas ∂V basically means "one thin sliver of the surface"; does this relate to the difference between ∂ andd?)—Ben FrantzDale23:29, 18 May 2006 (UTC)[reply]

In topology, ∂S denotes theboundary of a surfaceS. As far as I know, this is unrelated to the symbol's use to denote a partial differential (I've only just started seeing this formally, so I may be wrong). I also read this on an article here on Wikipedia which was about the different uses of the symbol; an article that I can't seem to locate now. If anyone can find it, I'd like to know. Anyway, I think this makes sense since the boundary of a volumeis a surface, as the boundary of an area is a path (at least intuitively). Hope this helps.Commander Nemet23:25, 31 May 2006 (UTC)[reply]

Since${\displaystyle F^{\prime }(P)=\frac{dF(P)}{dP}}$ and (with respect to P)${\displaystyle F^{\prime }(P,Q)=\frac{\mathrm{\partial }F(P,Q)}{\mathrm{\partial }P}}$, doesn't it follow that${\displaystyle F(P_d,Q)-F(P_b,Q)={\int }_{P_b}^{P_d}F(P,Q)\mathrm{\partial }P}$?  ~Kaimbridge~18:44, 20 May 2006 (UTC)[reply]

Yes, that is true, assuming that by the integrand you really meant${\displaystyle F_P(P,Q)}$ instead ofF itself. Also, I believe that the differential here would be justdP, not${\displaystyle \mathrm{\partial }P}$, though I can't give you a good reason why. What you would get after this integration is a function ofQ, which, if you were doing a double integral, would be integrated in the next step. I always thought of the double (or triple, etc.) integral as the "equivalent" of the partial derivative.Commander Nemet03:09, 1 June 2006 (UTC)[reply]

Actually, Imeant${\displaystyle F^{\prime }(P,Q)}$. In terms of full notation, should it be

${\displaystyle F(P_d,Q)-F(P_b,Q)={\int }_{P_b}^{P_d}{F}^{\prime }(P,Q)\mathrm{\partial }P={\int }_{P_b}^{P_d}{G}_p(P,Q)\mathrm{\partial }P}$, or should${\displaystyle F^{\prime }(P,Q)}$ be${\displaystyle F_p^{\prime }(P,Q)}$, with a subscript, too? As someone asked above in "Comparison with "d" notation", if you have "dP", you know it's regarding P, not Q, so what does${\displaystyle \mathrm{\partial }P}$ add? You could just as easily say

${\displaystyle E^{\prime }(P)=F^{\prime }(P,Q)=\sin (P)\cos (Q)}$, in which case

${\displaystyle E^{\prime }(P)dP=F^{\prime }(P,Q)\mathrm{\partial }P}$, right? In terms of it being a "double integral", a double integral integrates one variable, then the other. In the example I give above, only "'P'" is integrated.  ~Kaimbridge~19:28, 4 June 2006 (UTC)[reply]

In regards to your first question, the notation that I learned for partial derivatives includes, in this case,${\displaystyle \begin{array}{c}\frac{\mathrm{\partial }F(P,Q)}{\mathrm{\partial }P}\end{array}}$,${\displaystyle F_P(P,Q)}$, and${\displaystyle {\mathrm{\partial }}_{P}F(P,Q)}$.${\displaystyle F^{\prime }(P,Q)}$ is not used because it does not indicate which variable the derivative is taken with respect to.

In regards to your second question, I'm not really sure what you're asking, but I can tell you that, as far as I know, the partial differential symbol is used only in derivatives, so in the integral, for example, it would be justdP. I think the reason for this is that a differential is only "partial" in the context of a derivative. To me, at least, having a differential be "partial" otherwise would be meaningless.Commander Nemet22:49, 5 June 2006 (UTC)[reply]

It seems that 68.103.26.177 has changed the introduction to say the symbol is from the Greek instead of Cyrillic alphabet. I'm trying not to be hasty and call it vandalism, but unless there is some deeper meaning I am unaware of, the partial differential symbol looks a heck of a lot more like the cursivede (Cyrillic) than a Greekdelta. Furthermore, the intro now purports to say that the Greek letter in question is actually a "d"—something that doesn't exist in that alphabet.Commander Nemet23:25, 31 May 2006 (UTC)[reply]

• I agree with you that this is just wrong, if not vandalism. I therefore reverted it to the correct version.PanchoS00:29, 29 June 2006 (UTC)[reply]

• I think the firs person to use that notation was leibniz

According to Cajori'sHistory of Methematical Notations v. 2, p. 220 Leibniz used a 'δ' for the partial derivative.Thomasmeeks17:31, 11 December 2006 (UTC)[reply]

Many other mathematics-related articles have an implementation of the topic in various programming languages. Will anybody set up this page with such an example? --kanzure16:41, 19 July 2006 (UTC)[reply]

If so, please cite here. I'm doubtful based on # of hits for Google of:

• "partial derivative" Greek
• "partial derivative" Cyrillic

Earlier notation used Greek small delta δ, which looks like∂ than the Cyrillic Б. If no one can give an authoritative cite for Cyrillic, my vote would be for deleting sentence so suggesting it. Greek origin looks more plausible.Thomasmeeks02:25, 28 November 2006 (UTC)[reply]

The answer to the question in this section title seems to be in the following classic article:

Florian Cajori, "The History of Notations of the Calculus."Annals of Mathematics, 2nd Ser., Vol. 25, No. 1 (Sep., 1923), pp. 1-46

accessible throughJSTOR from many colleges and universities.Thomasmeeks01:08, 29 November 2006 (UTC)[reply]

Should the reference to the Cyrillic letter in the lead be deleted? For now, I believe so.Florian Cajori'sA History of Mathematical Notations (1928-29), still widely regarded as definitive for the period through its writing, makes no reference to the cursive Cyrillic letterde in the section on partial derivatives (v. 2 pp. 220-42). He refers only to the "rounded letter" after referencing earlier partial notations, including the 'ό' and 'd'. No other scholarly sources in the math reference section or the relevant sections of book collections on math history at a good university library make the Cyrillic connection. Nor does a search of Google Scholar give any primary source for: Cyrillic "partial derivative." So, I don't believe that the Cyrillic reference meets theWikipedia: verifiability requirement.Thomasmeeks 18:50, 11 December 2006 (UTC) ('rounded' for 'curved' editThomasmeeks 20:16, 11 December 2006 (UTC))Thomasmeeks20:18, 12 December 2006 (UTC)[reply]

Is there a WP policy that discourages arguments based on the number of Google 'hits'? If not, can someone please propose one?Here for example:From searching Google today:

• "partial derivative" American ~54000
• "partial derivative" English ~36400
• "partial derivative" French ~24000
• "partial derivative" Greek ~12300
• "partial derivative" Cyrillic ~519

On this basis we'd conclude that: Americans invented partial derivatives, and they probably spoke (a form of) English; their invention was aided by the French and quickly adopted by the Greek; and Russians generally know nothing about partial derivatives.
But let's also search >"partial derivative" Amrica<. It returns ~35800 'hits'.At least, Google claims that number of hits. It can't beverified. Disagree? Okay, then tell me what hit number 10000 is.
And what about other search engines???
—DIV (128.250.204.11801:34, 15 November 2007 (UTC))[reply]

Deletion from the previous Edit of the article of:

'∂' also corresponds to the small Greek delta 'ό' which was also used before the 20th century as partial derivative notation.

The problem is that that sentence follows a relatedparenthetical comment. In retrospect (I introduced the second sentence), if the sentence is going to be included at all, there is a case that both should be parenthetical (to avoid breaking the substanive exposition) or neither (possibly as a footnote or short separate section on the origin of the notation).Thomasmeeks14:56, 29 November 2006 (UTC)[reply]

"a partial derivative of a function of several variables is its derivative with respect to one of those variables with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary"

the total derivative is, in fact, the sum of the partial derivatives. i seem to remember explaining this before. but no matter. keep removing other people's contributions if you dont understand them—Precedingunsigned comment added by212.159.75.167 (talk)20:26, 13 October 2007 (UTC)[reply]

I disagree, the total derivative is not the sum of the partials. It is possible for all the partial derivatives to exist and yet the function not even be continuous (let alone have a total derivative). This is mentioned in the section *Formal Definition* : "even if all partial derivatives ∂f/∂ai(a) exist at a given point a, the function need not be continuous there." Brian Maurizi72.51.124.202 (talk)19:20, 21 April 2010 (UTC)[reply]

We should also discuss the numericalization in this article. At least illustrate how to numericalize the 2nd derivative. anyone agree?Jackzhp (talk)15:22, 3 May 2008 (UTC)[reply]

Hi, i'm currently studying on this topic(partial derivative) and i need some help to solve these problems.

1. The variable x and r are defined as

x = r cos θ and y = r sin θ.

a)Explain what is meant by i)( ∂y/∂r)θ and ii)( ∂y/∂r)x .b)Determine the values of (i).c)Determine value of (ii) by eliminating θ in the expressions for x and y.d)Determine the value of (ii) by the method of function of a function.e)Determine the value of :( ∂y/∂r)θ .( ∂y/∂r)x using method of b) and c).

2. If u = ax +by and v = bx – ay, determine the value of:

               (∂u/∂x)y.( ∂x/∂u)v

3. A, B and C are the angles of a triangle and u is a function defined as

                     u =  sin A sin B sin C.

i)Express u in terms of A and B only.ii)Prove that u is a maximum if the triangle is equilateral.

4. A vessel to be constructed is in the shape of a cylinder of radius r with equal conical ends, the semi-vertical angle of each cone being α. If V is the volume and S is the total surface area, show that

            S = 2V/r  + 2πr²(cosecα  - 2/3cotα )

If both r and α can vary, show that for a vessel of fixed volume and minimum surface area, cos α = 2/3, and determine r in terms of V.

Is there anyone able to enlighten me on this problems?THANKS a million...—Precedingunsigned comment added byXpsky (talk •contribs)05:35, 2 February 2009 (UTC)[reply]

Please ask this question atWikipedia:Reference desk/Mathematics. You will most certainly get a response, if you further add your input to the answering of these questions. In general, the talk page is not the most appropriate place to post questions. Further, the problems you quote are rather simple in nature, and similar worked examples can be found in a basic textbook on multivariable calculus. --PST04:38, 22 May 2009 (UTC)[reply]

Consideringtotal derivatives article, I feel that the equations should be

${\displaystyle \frac{\mathrm{d}V}{\mathrm{d}r}=\stackrel{\frac{\mathrm{\partial }V}{\mathrm{\partial }r}}{\overbrace{\frac{2\pi rh}{3}}}+\stackrel{\frac{\mathrm{\partial }V}{\mathrm{\partial }h}}{\overbrace{\frac{\pi r^2}{3}}}\frac{dh}{dr}}$

${\displaystyle \frac{\mathrm{d}V}{\mathrm{d}h}=\stackrel{\frac{\mathrm{\partial }V}{\mathrm{\partial }h}}{\overbrace{\frac{\pi r^2}{3}}}+\stackrel{\frac{\mathrm{\partial }V}{\mathrm{\partial }r}}{\overbrace{\frac{2\pi rh}{3}}}\frac{dr}{dh}}$

${\displaystyle k=\frac{h}{r}=\frac{dh}{dr}}$

instead of

${\displaystyle \frac{\mathrm{d}V}{\mathrm{d}r}=\stackrel{\frac{\mathrm{\partial }V}{\mathrm{\partial }r}}{\overbrace{\frac{2\pi rh}{3}}}+\stackrel{\frac{\mathrm{\partial }V}{\mathrm{\partial }h}}{\overbrace{\frac{\pi r^2}{3}}}\frac{\mathrm{\partial }h}{\mathrm{\partial }r}}$

${\displaystyle \frac{\mathrm{d}V}{\mathrm{d}h}=\stackrel{\frac{\mathrm{\partial }V}{\mathrm{\partial }h}}{\overbrace{\frac{\pi r^2}{3}}}+\stackrel{\frac{\mathrm{\partial }V}{\mathrm{\partial }r}}{\overbrace{\frac{2\pi rh}{3}}}\frac{\mathrm{\partial }r}{\mathrm{\partial }h}}$

${\displaystyle k=\frac{h}{r}=\frac{\mathrm{\partial }h}{\mathrm{\partial }r}}$

Please verify this.—Precedingunsigned comment added byLandroni (talk •contribs)15:00, 18 February 2009 (UTC)[reply]

Yes, you're right. I have corrected it, and removed the attention box.RupertMillard(Talk)10:04, 25 February 2009 (UTC)[reply]

The Introduction chapter mentions "to find the tangent line of the above function at (1, 1, 3)", as well as other instances of "(1, 1, 3)". But the function in question is f(x,y), which takes only two arguments. Its solution at (1, 1) is 3, though, as well asƒ(1, 1) = 3. Perhaps it is an error that has slipped by?

The question above was first posed by User:82.130.38.44, on 23:35, 19 April 2009 (UTC) but later removed by him. Perhaps did (s)he find a good explanation that (s)he can give to everybody? At this moment, I rather think (s)he was right to suspect an error. --Bdmy (talk)09:17, 20 April 2009 (UTC)[reply]

As 82.130.38.44 determined,ƒ(1, 1) = 3, so (1,1,3) are the (x,y,z) coordinates of a point on the surface. This is a good habit, as some functions will have multiple values of z for some x&y, in which case the derivatives at these different points will most probably be different.RupertMillard(Talk)09:57, 20 April 2009 (UTC)[reply]

Of course (1,1,3) is the point on the surface, but is it really a good way of phrasing things? You wouldn't talk of the derivative of the exponential function${\displaystyle e^x}$ at the point (1, e). Including the value as additional variable does not make any reasonable sense in my silly example, and I am not sure it does in this article (especially without a warning – for me it is just confusing). Also I don't agree with your point that some functions have multiple values at (x, y). For me they are just notfunctions in that case. You probably think about surfaces defined by an implicit equation, and that can be described by several "charts", involving several functions. --Bdmy (talk)10:25, 20 April 2009 (UTC)[reply]

This article discusses intuition but fails to discuss the main properties of the partial derivative; only at the end is the formal definition stated. For example, many facts found in a basic calculus textbook are not discussed, regarding the partial derivatives. What worries me, is that even the content of a basic calculus textbook is not adequate enough for this concept. --PST04:35, 22 May 2009 (UTC)[reply]

Doesn't f sub x = the partial of f / the partial of x? (not the partial of f / the partial of y)

see basic definition, where f sub a takes y to be the variable.

No, that's not what it means in that case. Read the defitions given just before that. The subscript doesn't represent any sort of derivative at all. That conflicts with the notation used elsewhere in the article, so that's potentially confusing. But the paragraph preceding that part is explicit about what the notation means.Michael Hardy (talk)22:36, 20 September 2009 (UTC)[reply]

My compliments to whoever has guided this wiki page, be it a group or an individual.

The introduction is EXCELLENT!— Precedingunsigned comment added by220.165.68.56 (talk)07:13, 8 January 2013 (UTC)[reply]

In theNotation-section it says:

${\displaystyle \frac{{\mathrm{\partial }}^{2}f}{\mathrm{\partial }y\mathrm{\partial }x}=...=f_{xy}}$

But in theHigher order partial derivatives-section it says:

${\displaystyle \frac{{\mathrm{\partial }}^{2}f}{\mathrm{\partial }x\mathrm{\partial }y}\equiv ...\equiv f_{xy}}$

So my basic question is the meaning of${\displaystyle \frac{{\mathrm{\partial }}^{2}f}{\mathrm{\partial }x\mathrm{\partial }y}}$:

1. Does it mean, that you first differentiate with respect tox and then with respect toy, because that is the order of the denominator?
2. Or does it mean, that you first differentiate with respect toy and then with respect tox, because derivative operators always work on the stuff, that comes after it? (So basically read from right to left, rather from left to right?)134.60.31.73 (talk)16:27, 10 April 2015 (UTC)[reply]

The theorem onsymmetry of second derivatives says that the two are equal--you get the same thing regardless of the order in which you take the derivatives. (This is mentioned at the end of the subsection "Formal definition".) So it doesn't matter how you interpret the notation, though it is standard to mean${\displaystyle \frac{{\mathrm{\partial }}^{2}f}{\mathrm{\partial }x\mathrm{\partial }y}\equiv \mathrm{\partial }\left(\frac{\mathrm{\partial }f}{\mathrm{\partial }x}\right)/\mathrm{\partial }y,}$ as is implicit in the H matrix in the articleHessian matrix.Loraof (talk)20:09, 9 May 2015 (UTC)[reply]

Thank you for your answer. But in the same articlesymmetry of second derivatives is says that this symmetry does only hold under special requirements, so it is not generally true, see here [[1]]. Soin the general case it does matter how I interpret the notation. Also in your mentioned articleHessian matrix I cannot find any passage that supports your mentioned "standard", perhaps you can elaborate? I would have chosen the opposite interpretation (my second alternative from my initial post).134.60.31.73 (talk)12:06, 19 June 2015 (UTC)[reply]

The comment(s) below were originally left atTalk:Partial derivative/CommentsTalk:Partial derivative/Comments, and are posted here for posterity. Followingseveral discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Last edited at 20:47, 20 September 2009 (UTC).Substituted at 02:27, 5 May 2016 (UTC)

I think the red term in the quote

The partial derivative of a function${\displaystyle f(x,y,\dots )}$ with respect to the variable${\displaystyle x}$ is variously denoted by${\displaystyle f_x^{\mathrm{\prime }},f_x,{\mathrm{\partial }}_{x}f,D_{x}f,D_{1}f,...}$

can not be correct, can it? I think that would be the undifferentiated function, not the derivative. --Yukterez (talk)02:01, 4 September 2017 (UTC)[reply]

It seems that there are non-trivial aspects of the use of partial derivativesin the context of math-based sciences akaapplied mathematics. An example is one connected to theGibbs-Duhem equation. I also notice some objection to the expressionmath-based sciences recently removed byuser:Deacon Vorbis.--82.137.15.34 (talk)11:35, 25 October 2017 (UTC)[reply]

You haven't asked a question or made a proposal or anything, but I'll add that essentially all science is "math-based". --Deacon Vorbis (talk)14:31, 25 October 2017 (UTC)[reply]

True (almost), but there are some areas from particular sciences which are more qualitative-descriptive.--82.137.11.191 (talk)21:05, 25 October 2017 (UTC)[reply]

A very interesting area or (applied) science (that is) rather qualitative (math is not required, at least as I know, in the educational system preparing those professionals) is medicine! (Thoughts?)--82.137.15.37 (talk)00:12, 26 October 2017 (UTC)[reply]

IP, i think the issue with your addition is that there are many, many applications in the sciences. In medical R&D partial derivatives are important in modelling blood flow (imp for cardiology, medical device development), pharmacology, medical imaging, nuclear medicine. But not doctors so much. Radiologists, when determining what kind of scan to give someone maybe, but they mostly have software that does this for them.Jytdog (talk)01:30, 26 October 2017 (UTC)[reply]

Very intriguing and/or paradoxical the situation you mention about applications of math in medical and physiological context, but not directly influencing the formation/education of medical professionals involved informulating medical diagnoses. Perhaps this has to do with a misconception of a historical nature of bad legacy having its origin in Medieval times (before theScientific Revolution) when math was not very appreciated or developed in those medieval times when only the literary and rethorical aspects from theClassics thinkers of Antiquity were highly esteemed and medicine was seen as craft or practical art in contrast with theliberal arts.--82.137.11.35 (talk)11:40, 26 October 2017 (UTC)[reply]

It also seems that medical doctors feel aless stringent necessity touse and develop abilities/skills involving mathematical thinking andmathematical modelling applied in their activities. The mentioned applications with medical impact that use mathematical notions are therefore left to be developed by other categories of professionals who haveanother conceptual perspective and a betterappreciation of the importance of some math notions such as engineers. After all, the above mentioned applications can be envisioned to belong more in thebiomedical engineering perspective.--82.137.11.35 (talk)11:53, 26 October 2017 (UTC)[reply]

I suspect doctors not taking advanced math has more to do with simply needing to focus directly on the practice of medicine which doesn't require Calculus directly. Only the research aspects and Biomedical engineering aspects require it directly. I know several doctors have taken Differential equations, but it's generally not required to get into medical school.Nickalh (talk)07:26, 24 January 2024 (UTC)[reply]

The expression for the extensiveGibbs free energy G of a mixture is a sum of weighted partial derivatives, weights being the molar amounts of the components n_i.

${\displaystyle G=\sum _{i=1}^m{n}_i\overline{G_i},}$

where${\displaystyle \overline{G_i}}$ is the partial molar${\displaystyle G}$ of component${\displaystyle i}$ defined as:

${\displaystyle \overline{G_i}={\left(\frac{\mathrm{\partial }G}{\mathrm{\partial }{n}_i}\right)}_{T,P,n_{j\ne i},(\frac{x_1}{x_3})}}$

The first additional relation can be expressed equivalently in term oftotal molar Gibbs energy of the mixture${\displaystyle \tilde{G}}$ usingmole fractions x_i instead of n_i where x_i is obtained by diving each weight (called mole fractions) by the sum of all n_i in the system and using the property that the sum of x_i equals 1.

${\displaystyle \tilde{G}=\sum _{i=1}^m{x}_i\overline{G_i},}$

The weights x_i can be represented geometrically/graphically in case of ternary and quaternary systems usingternary plot and quaternary plot in equilateral triangle and tetrahedron, etc.)--82.137.9.233 (talk)15:45, 26 October 2017 (UTC)[reply]

The last relation can have the derivative taken at with respect to one of the weights, say x₂ at a constant ratio of the other 2 weights x₁ and x₃ because when the variable x₂ varies, only the sum of the other mole fractions can vary:--82.137.9.233 (talk)16:07, 26 October 2017 (UTC)[reply]

${\displaystyle \tilde{G}=\sum _{i=1}^m{x}_i\overline{G_i},}$

From the sum of mole fractions equals 1 follows that:

${\displaystyle d{x}_3=-(d{x}_1+d{x}_2)}$

${\displaystyle {\left(\frac{\mathrm{\partial }{x}_1}{\mathrm{\partial }{x}_2}\right)}_{\frac{x_1}{x_3}}=-\frac{x_1}{1-x_2}}$--82.137.12.199 (talk)00:25, 27 October 2017 (UTC)[reply]

This sentence seems confusing to me since the quotient which is abovedoes not contain x_i per se.

(Incidentally, I notice that the mathematical font used for equationsis different from that used for inline formulas.)— Precedingunsigned comment added by130.79.10.22 (talk)13:02, 16 January 2019 (UTC)[reply]

The formula doesn't contain${\displaystyle x_i}$ because this variable has been particularized to the 'specific values'${\displaystyle a_i}$ and${\displaystyle a_i+h}$. It is referring to the variable of the function${\displaystyle f(x_1,\dots ,x_i,\dots ,x_n)}$ one line before the formula in question.

Regarding the font, yes, that is a common problem in Wikipedia. Since it is a collaboration it happens that content added by different people have different styles. Anyone can fix problems like that. Including you if you like.Cactus0192837465 (talk)15:19, 16 January 2019 (UTC)[reply]

Thanks for the feedback. I do not argue about the correctness of the definition. It is certainly very clear for someone who already understands the concept. My point is that I find that "particularization" not explicit enough. The French version seems to me to be a bit more explicit -- though still far from ideal for a beginner.130.79.10.22 (talk)13:57, 7 February 2019 (UTC)[reply]

Already in precalculus, students learn the difference between a function${\displaystyle f}$ and its value${\displaystyle f(x,y)}$ for a particular argument${\displaystyle (x,y)}$.In this article, this distinction seems to be forgotten, leading to meaningless expressions. In particular, if${\displaystyle f}$ is a function, then${\displaystyle \frac{\mathrm{\partial }f}{\mathrm{\partial }x}}$ is meaningless since${\displaystyle f}$ is unrelated to any particular choice of variable names. Proper forms are${\displaystyle \frac{\mathrm{\partial }f(x,y)}{\mathrm{\partial }x}}$ or${\displaystyle \frac{\mathrm{\partial }u}{\mathrm{\partial }x}}$ where${\displaystyle u=f(x,y)}$. The aforementioned confusion causes many problems, for instance, when changing variables or formulating the chain rule. For various reasons, I prefer leaving it to the editors to make all necessary changes.--Boute (talk)07:18, 2 October 2019 (UTC)[reply]

@Boute: Sorry, that is not correct or supported by usage in sources.${\displaystyle \frac{\mathrm{\partial }f}{\mathrm{\partial }x}}$ is understood to be another function wherever it is defined. After all, the partial derivative operator acts onfunctions and not their values. Since x and y will always be understood to be free parameters (pairs of which comprise the domain) here, the partial derivative function and the partial derivative functionevaluated at x, y are basically identical. --Jasper Deng(talk)07:27, 2 October 2019 (UTC)[reply]

@Jasper: All depends on the sources, many of which are the cause of the confusion. Up to (rather wide) variations in notation, there are essentially two kinds of partial derivative operators: (i) those operating on functions, say,${\displaystyle {\mathrm{D}}_{k}f}$ for the partial derivative of${\displaystyle f}$ with respect to the${\displaystyle k}$-th argument (and${\displaystyle {\mathrm{D}}_{k}f}$ is a function) and (ii) those operating on expressions denoting numbers, for instance${\displaystyle \frac{\mathrm{\partial }(x^2+xy)}{\mathrm{\partial }x}=2x+y}$. For a proper treatmentand in practical use, you need both. Since I dislike public discussions on the web with everyone barging in and expecting reactions, you can send me an e-mail and I send you a paper (in preparation) on this topic. My e-mail address is easy to find but I won't advertise it here.--Boute (talk)07:58, 2 October 2019 (UTC)[reply]

@Boute: We are not here toWP:RIGHTGREATWRONGS about everyone's use of notation, so any and all materials used to make your case must ultimately end up here for everyone to see as we use what sources use. That said, if we can demonstrate the existence of a strong consensus on this distinction made here, then we should have a remark on this in the notation section. Note though that apolynomial, by abuse of notation, is usually treated as the function defined by evaluating it at a given point. There are some implicit assumptions made: the domain consists of tuples only consisting of values of the specified variables and those specified variables are free of each other (otherwise either we have[2], or the joke doesn't work, respectively).--Jasper Deng(talk)08:16, 2 October 2019 (UTC)[reply]

@Jasper: I don't thinkWP:RIGHTGREATWRONGS applies here since saying 'that is incorrect' is not my way of clarifying issues. Some sources are self-consistent, others are not. This is a matter of logic, not notation. Whatever the notation, the distinction between a function${\displaystyle f}$ and the expression${\displaystyle f(x,y)}$ is essential, and so is the distinction between partial derivation operators mapping functions to functions and those mapping expressions to expressions. Since this distinction is conceptually clear, can there be any doubt about consensus? Consensus about notation is less important: anyone can use a preferred variant, provided the concepts are not confused. Aside: the joke about${\displaystyle {\mathrm{e}}^x}$ works perfectly well since${\displaystyle {\mathrm{e}}^x}$ is an expression in the variable${\displaystyle x}$, not a function, so${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }x}{\mathrm{e}}^x={\mathrm{e}}^x}$ and${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }y}{\mathrm{e}}^x=0}$, which precisely demonstrates the difference with partial derivation operators mapping functions to functions. Allow me to repeat that discussion on the web is totally unsuitable for resolving anything since it encourages reacting too quickly instead of thinking things through and editors seem to prefer unreliable sources over logic. Feel free to contact me as suggested earlier, but please understand that I cannot spend time on these pages and henceforth will react only to e-mail..--Boute (talk)13:27, 2 October 2019 (UTC)[reply]

Prompted by the preceding discussion, I compared about a dozen accounts of partial derivatives in different textbooks. Although 100 sloppy sources don't outweigh a single careful one, I was aghast at how many sources are sloppy, so the confusion is understandable. The most careful treatment of partial derivatives, including comparison between notations, was found inMathematical Analysis by Thomas M. Flett (1966, before math reforms caused a dip in mathematical diligence). By coincidence (which I won't even called fortunate since variants, when used properly, work equally well), Flett's conventions are the same as the ones I mentioned earlier.

a. On page 351 (bottom), assuming${\displaystyle f}$ is a function from a subset of${\displaystyle {\mathbb{R}}^n}$ to${\displaystyle \mathbb{R}}$, Flett defines${\displaystyle {\mathrm{D}}_{j}f}$ as a real-valued function called thepartial derivative of${\displaystyle f}$ with respect to the${\displaystyle j}$-th coordinate.

b. On page 352 (top), Flett statesWe also write${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }{x}_j}(f(x_1,\dots ,x_n))}$ for the partial derivative of${\displaystyle f}$ with respect to the${\displaystyle j}$-th coordinate at the point${\displaystyle (x_1,\dots ,x_n)}$ and mentions${\displaystyle \frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_j}}$ as an 'other notation' (which he never uses) whose meaning depends on the context (a polite way of saying that it is inherently ambiguous).

c.On page 353, a few examples illustrate how both${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }{x}_j}}$ and${\displaystyle {\mathrm{D}}_j}$ are used.

If someone can tell me how to attach documents to Wikipedia `Talk' items, I can do so for copies of the three relevant pages from Flett's book. --Boute (talk)17:06, 2 October 2019 (UTC)[reply]

Okay, but there is one and only one definition of a "partial derivative", which is as anoperator on a function. Treating an expression as the function defined by it is a commonabuse of notation and we must implicitly choose such a function when applying a partial derivative operator. The point of my bringing up RIGHTGREATWRONGS isWP:DUE weight. This is what I mean by "consensus". You seem to have pointed out a great variety of treatments in textbooks but not any sources that specificallycritique all of these notations and the difference between them. In particular, the great majority of sources appear to agree that${\displaystyle \frac{\mathrm{\partial }f}{\mathrm{\partial }x}}$ is meaningful and a great deal of professors at my institution (one named in the2019 college admissions scandal) teach with a whole lot of these notations, including differential geometry classes.

You seem to imply that readers frequently get confused. However, certainly I have not (though I'm biased as I've formally instructed this content before).--18:48, 2 October 2019 (UTC)

I recommend that the definition of partial derivative either include a hyperlink to Standard Basis or define e_i.Cahdfin13773 (talk)00:25, 19 July 2023 (UTC)[reply]

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