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curl -X POST \
  'http://laksa018.int.ahrefs:8124/' \
  -H 'Content-Type: text/plain' \
  -H 'X-ClickHouse-Database: crawler3' \
  -H 'Authorization: Basic YXBpOg==' \
  -d 'SELECT getAhrefsURLFromUnparsed(src_unparsed) AS found_url, ifNull(toUnixTimestamp(download_stamp), 0) AS crawl_time, ifNull(toUnixTimestamp(props_url_first_seen), 0) AS first_indexed_time, download_http_code AS http_code, src_unparsed AS src_unparsed, src_root_hash AS src_root_hash, history_drop_reason AS history_drop_reason, meta_title AS meta_title, meta_descriptions AS meta_descriptions, attrs_boilerpipe_text AS attrs_boilerpipe_text, attrs_markdown AS attrs_markdown, attrs_readable_markdown AS attrs_readable_markdown, meta_canonical AS meta_canonical FROM crawler3.page_info_local FINAL PREWHERE (src_root_hash, src_unparsed) IN ((getAhrefsRootHashFromUnparsed(getAhrefsUnparsedNoserviceFromURL(\'https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation\')), getAhrefsUnparsedNoserviceFromURL(\'https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation\'))) FORMAT JSONEachRow'

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{"found_url":"https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation","crawl_time":1773574772,"first_indexed_time":0,"http_code":200,"src_unparsed":"com,stackexchange!math,\/questions\/127361\/second-order-partial-derivatives-notation s443","src_root_hash":"8045678284012640218","history_drop_reason":null,"meta_title":"calculus - Second order partial derivatives - notation - Mathematics Stack Exchange","meta_descriptions":[],"attrs_boilerpipe_text":"4\n$\\begingroup$\nI have seen both of these used, and people around me seem to disagree, so which one is correct: (first derivative with respect to x, then y):\n(1) $$\\frac{\\partial }{\\partial y}(\\frac{\\partial f}{\\partial x}) = \\frac{\\partial^{2} f}{\\partial x\\partial y}$$\n(2) $$\\frac{\\partial }{\\partial y}(\\frac{\\partial f}{\\partial x}) = \\frac{\\partial^{2} f}{\\partial y\\partial x}$$\nand why? (reasons, history?)\nasked\nApr 2, 2012 at 19:52\n$\\endgroup$\n5\n+50\n$\\begingroup$\n$\\def\\part#1#2{{\\partial#1\\over\\partial#2}}$\n$\\def\\parts#1#2#3{{\\partial^2#1\\over\\partial#2\\,\\partial#3}}$\nOn the left hand side of your equations, you have the symbol \"$\\part{\\vphantom f}y\\bigl(\\part f x\\bigr)\"$.\nBy definition this  is the partial derivative of the function $\\part fx$ with respect to $y$. \nSo, upon encountering this symbol, you   take the function $\\part fx$ and  then take its partial with respect to $y$.  The natural notation of the type on the right hand side of your equations  is the notation used in (2) of your post:\n$$\\tag{3}\r\n\\part{\\vphantom f}y\\Bigl(\\part f x\\Bigr)=\\parts f y x.\r\n$$\nI will not surmise why this is the \"natural\" notation, but will point out that $(3)$ gives  the adopted definition for the symbol $\\parts f y x$ in\nany\ncalculus\/analysis text, or  any other \"credible\" source, you'll find.\nI emphasise here  that $(3)$\ndefines\nthe symbol $\\parts f y x$; that this sometimes gives an expression that equals $\\parts f x y$ is irrelevant. (Of course, for certain functions, what you wrote in (1) would be correct; but its correctness would follow from the result of a theorem, not from the definition of the symbols.)\nanswered\nApr 8, 2012 at 17:32\n$\\endgroup$\n2\n1\n$\\begingroup$\nThe order is important when the function is not $C^2$. That is, the second derivatives (in relation to any combination of two variables) of $f$ are continuous functions. If the function is $C^2$ then it doesn't matter the order in which the variables appear.\nThis is a widely known result called Schwarz's Theorem, but it seems that there are other names for it. Check out for more in\nhttp:\/\/en.wikipedia.org\/wiki\/Symmetry_of_partial_derivatives\nunder the \"Clairaut's theorem\" subtitle.\nanswered\nApr 2, 2012 at 20:03\n$\\endgroup$\n11\n0\n$\\begingroup$\nhere i uesd notation $f_{xy}$ and $f_{yx}$ for double derivative.\\\nThere is some theorem on equality of $f_{xy}$ and $f_{yx}$.\nTheorem : (Young's theorem)\\\nLet $f$ be a real function defined on non-empty open set $E$ subset of $R^{2}$. If $f_{x}$ and $f_{y}$ exist in some nbhd. of $(x,y)$ and are both differentiable at point $(x,y)$ with respect to $x$ and $y$ then $f_{xy}=f_{yx}$ at point $(x,y)$.\nTheorem: (Schwartz's theorem)\nLet $f:E\\to R$ be a function such that its partial derivatives $f_{x},f_{y}$ and $f_{xy}$ exist and are continuous in a nbhd. of a point $(x,y)$ then $f_{yx}$ exist such that $f_{xy}=f_{yx}$.\nNote: One can see that the converse of above theorem need not be true.\nanswered\nApr 8, 2012 at 8:51\n$\\endgroup$\n0\n$\\begingroup$\nI always saw the second : the last operation you made is derivating with respect Y and tha symbol appears in the same order .\nIn the other notation the order is inverse fxy it means the last operation you made is derivating with respect y.\nBut : if both fxy and fyx are defined at a neighborhood of a point and are continuous at the point their value is the same at the point.See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10\nanswered\nApr 8, 2012 at 10:32\n$\\endgroup$\n1\nYou must\nlog in\nto answer this question.\nStart asking to get answers\nFind the answer to your question by asking.\nAsk question\nExplore related questions\nSee similar questions with these tags.","attrs_markdown":"[Skip to main content](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation#content)\n\n#### Stack Exchange Network\nStack Exchange network consists of 183 Q\\&A communities including [Stack Overflow](https:\/\/stackoverflow.com\/), the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.\n\n[Visit Stack Exchange](https:\/\/stackexchange.com\/)\n\n1. - [Tour Start here for a quick overview of the site](https:\/\/math.stackexchange.com\/tour)\n   - [Help Center Detailed answers to any questions you might have](https:\/\/math.stackexchange.com\/help)\n   - [Meta Discuss the workings and policies of this site](https:\/\/math.meta.stackexchange.com\/)\n   - [About Us Learn more about Stack Overflow the company, and our products](https:\/\/stackoverflow.co\/)\n2. ### [current community](https:\/\/math.stackexchange.com\/)\n   - [Mathematics](https:\/\/math.stackexchange.com\/)\n     [help](https:\/\/math.stackexchange.com\/help) [chat](https:\/\/chat.stackexchange.com\/?tab=site&host=math.stackexchange.com)\n   - [Mathematics Meta](https:\/\/math.meta.stackexchange.com\/)\n   ### your communities\n   [Sign up](https:\/\/math.stackexchange.com\/users\/signup?ssrc=site_switcher&returnurl=https%3A%2F%2Fmath.stackexchange.com%2Fquestions%2F127361%2Fsecond-order-partial-derivatives-notation) or [log in](https:\/\/math.stackexchange.com\/users\/login?ssrc=site_switcher&returnurl=https%3A%2F%2Fmath.stackexchange.com%2Fquestions%2F127361%2Fsecond-order-partial-derivatives-notation) to customize your list.\n   \n   ### [more stack exchange communities](https:\/\/stackexchange.com\/sites)\n   [company blog](https:\/\/stackoverflow.blog\/)\n3. 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[Learn more](https:\/\/stackoverflow.co\/internal\/?utm_medium=referral&utm_source=math-community&utm_campaign=side-bar&utm_content=explore-teams-compact)\n\n**Stack Internal**\n\nKnowledge at work\n\nBring the best of human thought and AI automation together at your work.\n\n[Explore Stack Internal](https:\/\/stackoverflow.co\/internal\/?utm_medium=referral&utm_source=math-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover)\n\n# [Second order partial derivatives - notation](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation)\n[Ask Question](https:\/\/math.stackexchange.com\/questions\/ask)\n\nAsked\n\n13 years, 11 months ago\n\nModified [13 years, 11 months ago](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation?lastactivity \"2012-04-08 17:32:24Z\")\n\nViewed 5k times\n\n4\n\n\\$\\\\begingroup\\$\n\nI have seen both of these used, and people around me seem to disagree, so which one is correct: (first derivative with respect to x, then y):\n\n(1) \\$\\$\\\\frac{\\\\partial }{\\\\partial y}(\\\\frac{\\\\partial f}{\\\\partial x}) = \\\\frac{\\\\partial^{2} f}{\\\\partial x\\\\partial y}\\$\\$\n\n(2) \\$\\$\\\\frac{\\\\partial }{\\\\partial y}(\\\\frac{\\\\partial f}{\\\\partial x}) = \\\\frac{\\\\partial^{2} f}{\\\\partial y\\\\partial x}\\$\\$\n\nand why? (reasons, history?)\n\n- [calculus](https:\/\/math.stackexchange.com\/questions\/tagged\/calculus \"show questions tagged 'calculus'\")\n- [multivariable-calculus](https:\/\/math.stackexchange.com\/questions\/tagged\/multivariable-calculus \"show questions tagged 'multivariable-calculus'\")\n- [notation](https:\/\/math.stackexchange.com\/questions\/tagged\/notation \"show questions tagged 'notation'\")\n- [derivatives](https:\/\/math.stackexchange.com\/questions\/tagged\/derivatives \"show questions tagged 'derivatives'\")\n\n[Share](https:\/\/math.stackexchange.com\/q\/127361 \"Short permalink to this question\")\n\nCite\n\nFollow\n\n[edited Apr 4, 2012 at 1:35](https:\/\/math.stackexchange.com\/posts\/127361\/revisions \"show all edits to this post\")\n\nasked Apr 2, 2012 at 19:52\n\n[![Thomas's user avatar](https:\/\/www.gravatar.com\/avatar\/e55135a3fbd2d22f1bd86f34cec13d3c?s=64&d=identicon&r=PG)](https:\/\/math.stackexchange.com\/users\/26188\/thomas)\n\n[Thomas](https:\/\/math.stackexchange.com\/users\/26188\/thomas)\n\n44\\.9k1313 gold badges7777 silver badges140140 bronze badges\n\n\\$\\\\endgroup\\$\n\n[Add a comment](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation \"Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.\")   \\|\n\n## 4 Answers 4\nSorted by:\n\n[Reset to default](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation?answertab=scoredesc#tab-top)\n\n5\n\n\\+50\n\n\\$\\\\begingroup\\$\n\n\\$\\\\def\\\\part\\#1\\#2{{\\\\partial\\#1\\\\over\\\\partial\\#2}}\\$ \\$\\\\def\\\\parts\\#1\\#2\\#3{{\\\\partial^2\\#1\\\\over\\\\partial\\#2\\\\,\\\\partial\\#3}}\\$\n\nOn the left hand side of your equations, you have the symbol \"\\$\\\\part{\\\\vphantom f}y\\\\bigl(\\\\part f x\\\\bigr)\"\\$. By definition this is the partial derivative of the function \\$\\\\part fx\\$ with respect to \\$y\\$. So, upon encountering this symbol, you take the function \\$\\\\part fx\\$ and then take its partial with respect to \\$y\\$. The natural notation of the type on the right hand side of your equations is the notation used in (2) of your post: \\$\\$\\\\tag{3} \\\\part{\\\\vphantom f}y\\\\Bigl(\\\\part f x\\\\Bigr)=\\\\parts f y x. \\$\\$\n\nI will not surmise why this is the \"natural\" notation, but will point out that \\$(3)\\$ gives the adopted definition for the symbol \\$\\\\parts f y x\\$ in *any* calculus\/analysis text, or any other \"credible\" source, you'll find.\n\nI emphasise here that \\$(3)\\$ *defines* the symbol \\$\\\\parts f y x\\$; that this sometimes gives an expression that equals \\$\\\\parts f x y\\$ is irrelevant. (Of course, for certain functions, what you wrote in (1) would be correct; but its correctness would follow from the result of a theorem, not from the definition of the symbols.)\n\n[Share](https:\/\/math.stackexchange.com\/a\/129358 \"Short permalink to this answer\")\n\nCite\n\nFollow\n\nanswered Apr 8, 2012 at 17:32\n\n[![David Mitra's user avatar](https:\/\/i.sstatic.net\/vJm8f.png?s=64)](https:\/\/math.stackexchange.com\/users\/18986\/david-mitra)\n\n[David Mitra](https:\/\/math.stackexchange.com\/users\/18986\/david-mitra)\n\n76\\.8k1111 gold badges149149 silver badges205205 bronze badges\n\n\\$\\\\endgroup\\$\n\n2\n\n- \\$\\\\begingroup\\$ Ok, so I think that this was the answer that I was looking for. I had seen the other notation in a business calculus book and they were being pretty consistent about it. I had also asked a Ph.D. student (with specialty in analysis) and he said that he used the first (1) notation. Can you think of any reason why someone might want the other notation? (option (1)) \\$\\\\endgroup\\$\n  \n  Thomas\n  \n  – [Thomas](https:\/\/math.stackexchange.com\/users\/26188\/thomas \"44,901 reputation\")\n  \n  2012-04-08 23:21:11 +00:00\n  \n  Commented Apr 8, 2012 at 23:21\n- 1\n  \\$\\\\begingroup\\$ @Thomas Most, if not all, texts (including [Paul's online notes](http:\/\/tutorial.math.lamar.edu\/Classes\/CalcIII\/HighOrderPartialDerivs.aspx), Stewart's Calculus, Apostol's Calculus, etc...) I've seen use (2) as the definition. It does seem natural to me since you can \"multiply the left side\" to get the right side. I can think of no reason to prefer the other. In any case, the symbol does need a definition, as the two expressions can give different results for not so nice functions. For example, there are differentiable functions with unequal mixed derivatives. \\$\\\\endgroup\\$\n  \n  David Mitra\n  \n  – [David Mitra](https:\/\/math.stackexchange.com\/users\/18986\/david-mitra \"76,753 reputation\")\n  \n  2012-04-08 23:35:50 +00:00\n  \n  Commented Apr 8, 2012 at 23:35\n\n[Add a comment](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation \"Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.\")   \\|\n\n1\n\n\\$\\\\begingroup\\$\n\nThe order is important when the function is not \\$C^2\\$. That is, the second derivatives (in relation to any combination of two variables) of \\$f\\$ are continuous functions. If the function is \\$C^2\\$ then it doesn't matter the order in which the variables appear.\n\nThis is a widely known result called Schwarz's Theorem, but it seems that there are other names for it. Check out for more in <http:\/\/en.wikipedia.org\/wiki\/Symmetry_of_partial_derivatives> under the \"Clairaut's theorem\" subtitle.\n\n[Share](https:\/\/math.stackexchange.com\/a\/127369 \"Short permalink to this answer\")\n\nCite\n\nFollow\n\nanswered Apr 2, 2012 at 20:03\n\n[![Marra's user avatar](https:\/\/i.sstatic.net\/bjgAJ.jpg?s=64)](https:\/\/math.stackexchange.com\/users\/27071\/marra)\n\n[Marra](https:\/\/math.stackexchange.com\/users\/27071\/marra)\n\n5,0632929 silver badges6868 bronze badges\n\n\\$\\\\endgroup\\$\n\n11\n\n- \\$\\\\begingroup\\$ please explain all difference between all the cases and how come they are different? \\$\\\\endgroup\\$\n  \n  Victor\n  \n  – [Victor](https:\/\/math.stackexchange.com\/users\/12796\/victor \"8,616 reputation\")\n  \n  2012-04-02 20:04:28 +00:00\n  \n  Commented Apr 2, 2012 at 20:04\n- 1\n  \\$\\\\begingroup\\$ @Victor Have you taken a multi variable calculus course? If so, it should be clear that they will infact be different for some classes of functions and whatnot\\! \\$\\\\endgroup\\$\n  \n  user21436\n  \n  – user21436\n  \n  2012-04-02 20:08:00 +00:00\n  \n  Commented Apr 2, 2012 at 20:08\n- \\$\\\\begingroup\\$ @KannappanSampath - no, never taken it before. \\$\\\\endgroup\\$\n  \n  Victor\n  \n  – [Victor](https:\/\/math.stackexchange.com\/users\/12796\/victor \"8,616 reputation\")\n  \n  2012-04-02 20:10:09 +00:00\n  \n  Commented Apr 2, 2012 at 20:10\n- 1\n  \\$\\\\begingroup\\$ @Victor Then I suggest you do it. :P \\$\\\\endgroup\\$\n  \n  user21436\n  \n  – user21436\n  \n  2012-04-02 20:10:47 +00:00\n  \n  Commented Apr 2, 2012 at 20:10\n- \\$\\\\begingroup\\$ Hi, thanks for the response, but I was asking about which notation is correct. \\$\\\\endgroup\\$\n  \n  Thomas\n  \n  – [Thomas](https:\/\/math.stackexchange.com\/users\/26188\/thomas \"44,901 reputation\")\n  \n  2012-04-02 20:17:15 +00:00\n  \n  Commented Apr 2, 2012 at 20:17\n\n\\|   [Show **6** more comments](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation \"Expand to show all comments on this post\")\n\n0\n\n\\$\\\\begingroup\\$\n\nhere i uesd notation \\$f\\_{xy}\\$ and \\$f\\_{yx}\\$ for double derivative.\\\\ There is some theorem on equality of \\$f\\_{xy}\\$ and \\$f\\_{yx}\\$. Theorem : (Young's theorem)\\\\ Let \\$f\\$ be a real function defined on non-empty open set \\$E\\$ subset of \\$R^{2}\\$. If \\$f\\_{x}\\$ and \\$f\\_{y}\\$ exist in some nbhd. of \\$(x,y)\\$ and are both differentiable at point \\$(x,y)\\$ with respect to \\$x\\$ and \\$y\\$ then \\$f\\_{xy}=f\\_{yx}\\$ at point \\$(x,y)\\$.\n\nTheorem: (Schwartz's theorem) Let \\$f:E\\\\to R\\$ be a function such that its partial derivatives \\$f\\_{x},f\\_{y}\\$ and \\$f\\_{xy}\\$ exist and are continuous in a nbhd. of a point \\$(x,y)\\$ then \\$f\\_{yx}\\$ exist such that \\$f\\_{xy}=f\\_{yx}\\$.\n\nNote: One can see that the converse of above theorem need not be true.\n\n[Share](https:\/\/math.stackexchange.com\/a\/129248 \"Short permalink to this answer\")\n\nCite\n\nFollow\n\nanswered Apr 8, 2012 at 8:51\n\n[![Kns's user avatar](https:\/\/www.gravatar.com\/avatar\/1579adfb80a078396b9688d8aa6b1039?s=64&d=identicon&r=PG)](https:\/\/math.stackexchange.com\/users\/27579\/kns)\n\n[Kns](https:\/\/math.stackexchange.com\/users\/27579\/kns)\n\n3,35322 gold badges3232 silver badges4343 bronze badges\n\n\\$\\\\endgroup\\$\n\n[Add a comment](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation \"Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.\")   \\|\n\n0\n\n\\$\\\\begingroup\\$\n\nI always saw the second : the last operation you made is derivating with respect Y and tha symbol appears in the same order . In the other notation the order is inverse fxy it means the last operation you made is derivating with respect y. But : if both fxy and fyx are defined at a neighborhood of a point and are continuous at the point their value is the same at the point.See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10\n\n[Share](https:\/\/math.stackexchange.com\/a\/129264 \"Short permalink to this answer\")\n\nCite\n\nFollow\n\n[edited Apr 8, 2012 at 11:01](https:\/\/math.stackexchange.com\/posts\/129264\/revisions \"show all edits to this post\")\n\nanswered Apr 8, 2012 at 10:32\n\n[![alpha.Debi's user avatar](https:\/\/www.gravatar.com\/avatar\/9863491b93c64d5849391d09bcb088f8?s=64&d=identicon&r=PG)](https:\/\/math.stackexchange.com\/users\/20863\/alpha-debi)\n\n[alpha.Debi](https:\/\/math.stackexchange.com\/users\/20863\/alpha-debi)\n\n1,09477 silver badges77 bronze badges\n\n\\$\\\\endgroup\\$\n\n1\n\n- \\$\\\\begingroup\\$ See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10 \\$\\\\endgroup\\$\n  \n  alpha.Debi\n  \n  – [alpha.Debi](https:\/\/math.stackexchange.com\/users\/20863\/alpha-debi \"1,094 reputation\")\n  \n  2012-04-08 11:01:03 +00:00\n  \n  Commented Apr 8, 2012 at 11:01\n\n[Add a comment](https:\/\/math.stackexchange.com\/questions\/127361\/second-order-partial-derivatives-notation \"Use comments to ask for more information or suggest improvements. 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(reasons, history?)\n\nasked Apr 2, 2012 at 19:52\n\n[![Thomas's user avatar](https:\/\/www.gravatar.com\/avatar\/e55135a3fbd2d22f1bd86f34cec13d3c?s=64&d=identicon&r=PG)](https:\/\/math.stackexchange.com\/users\/26188\/thomas)\n\n\\$\\\\endgroup\\$\n\n5\n\n\\+50\n\n\\$\\\\begingroup\\$\n\n\\$\\\\def\\\\part\\#1\\#2{{\\\\partial\\#1\\\\over\\\\partial\\#2}}\\$ \\$\\\\def\\\\parts\\#1\\#2\\#3{{\\\\partial^2\\#1\\\\over\\\\partial\\#2\\\\,\\\\partial\\#3}}\\$\n\nOn the left hand side of your equations, you have the symbol \"\\$\\\\part{\\\\vphantom f}y\\\\bigl(\\\\part f x\\\\bigr)\"\\$. By definition this is the partial derivative of the function \\$\\\\part fx\\$ with respect to \\$y\\$. So, upon encountering this symbol, you take the function \\$\\\\part fx\\$ and then take its partial with respect to \\$y\\$. The natural notation of the type on the right hand side of your equations is the notation used in (2) of your post: \\$\\$\\\\tag{3} \\\\part{\\\\vphantom f}y\\\\Bigl(\\\\part f x\\\\Bigr)=\\\\parts f y x. \\$\\$\n\nI will not surmise why this is the \"natural\" notation, but will point out that \\$(3)\\$ gives the adopted definition for the symbol \\$\\\\parts f y x\\$ in *any* calculus\/analysis text, or any other \"credible\" source, you'll find.\n\nI emphasise here that \\$(3)\\$ *defines* the symbol \\$\\\\parts f y x\\$; that this sometimes gives an expression that equals \\$\\\\parts f x y\\$ is irrelevant. (Of course, for certain functions, what you wrote in (1) would be correct; but its correctness would follow from the result of a theorem, not from the definition of the symbols.)\n\nanswered Apr 8, 2012 at 17:32\n\n[![David Mitra's user avatar](https:\/\/i.sstatic.net\/vJm8f.png?s=64)](https:\/\/math.stackexchange.com\/users\/18986\/david-mitra)\n\n\\$\\\\endgroup\\$\n\n2\n\n1\n\n\\$\\\\begingroup\\$\n\nThe order is important when the function is not \\$C^2\\$. That is, the second derivatives (in relation to any combination of two variables) of \\$f\\$ are continuous functions. If the function is \\$C^2\\$ then it doesn't matter the order in which the variables appear.\n\nThis is a widely known result called Schwarz's Theorem, but it seems that there are other names for it. Check out for more in <http:\/\/en.wikipedia.org\/wiki\/Symmetry_of_partial_derivatives> under the \"Clairaut's theorem\" subtitle.\n\nanswered Apr 2, 2012 at 20:03\n\n[![Marra's user avatar](https:\/\/i.sstatic.net\/bjgAJ.jpg?s=64)](https:\/\/math.stackexchange.com\/users\/27071\/marra)\n\n\\$\\\\endgroup\\$\n\n11\n\n0\n\n\\$\\\\begingroup\\$\n\nhere i uesd notation \\$f\\_{xy}\\$ and \\$f\\_{yx}\\$ for double derivative.\\\\ There is some theorem on equality of \\$f\\_{xy}\\$ and \\$f\\_{yx}\\$. Theorem : (Young's theorem)\\\\ Let \\$f\\$ be a real function defined on non-empty open set \\$E\\$ subset of \\$R^{2}\\$. If \\$f\\_{x}\\$ and \\$f\\_{y}\\$ exist in some nbhd. of \\$(x,y)\\$ and are both differentiable at point \\$(x,y)\\$ with respect to \\$x\\$ and \\$y\\$ then \\$f\\_{xy}=f\\_{yx}\\$ at point \\$(x,y)\\$.\n\nTheorem: (Schwartz's theorem) Let \\$f:E\\\\to R\\$ be a function such that its partial derivatives \\$f\\_{x},f\\_{y}\\$ and \\$f\\_{xy}\\$ exist and are continuous in a nbhd. of a point \\$(x,y)\\$ then \\$f\\_{yx}\\$ exist such that \\$f\\_{xy}=f\\_{yx}\\$.\n\nNote: One can see that the converse of above theorem need not be true.\n\nanswered Apr 8, 2012 at 8:51\n\n[![Kns's user avatar](https:\/\/www.gravatar.com\/avatar\/1579adfb80a078396b9688d8aa6b1039?s=64&d=identicon&r=PG)](https:\/\/math.stackexchange.com\/users\/27579\/kns)\n\n\\$\\\\endgroup\\$\n\n0\n\n\\$\\\\begingroup\\$\n\nI always saw the second : the last operation you made is derivating with respect Y and tha symbol appears in the same order . In the other notation the order is inverse fxy it means the last operation you made is derivating with respect y. But : if both fxy and fyx are defined at a neighborhood of a point and are continuous at the point their value is the same at the point.See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10\n\nanswered Apr 8, 2012 at 10:32\n\n[![alpha.Debi's user avatar](https:\/\/www.gravatar.com\/avatar\/9863491b93c64d5849391d09bcb088f8?s=64&d=identicon&r=PG)](https:\/\/math.stackexchange.com\/users\/20863\/alpha-debi)\n\n\\$\\\\endgroup\\$\n\n1\n## You must [log in](https:\/\/math.stackexchange.com\/users\/login?ssrc=question_page&returnurl=https%3A%2F%2Fmath.stackexchange.com%2Fquestions%2F127361) to answer this question.\nStart asking to get answers\n\nFind the answer to your question by asking.\n\n[Ask question](https:\/\/math.stackexchange.com\/questions\/ask)\n\nExplore related questions\n\nSee similar questions with these tags.","meta_canonical":null}

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Boilerpipe Text	4 $\begingroup$ I have seen both of these used, and people around me seem to disagree, so which one is correct: (first derivative with respect to x, then y): (1) $$\frac{\partial }{\partial y}(\frac{\partial f}{\partial x}) = \frac{\partial^{2} f}{\partial x\partial y}$$ (2) $$\frac{\partial }{\partial y}(\frac{\partial f}{\partial x}) = \frac{\partial^{2} f}{\partial y\partial x}$$ and why? (reasons, history?) asked Apr 2, 2012 at 19:52 $\endgroup$ 5 +50 $\begingroup$ $\def\part#1#2{{\partial#1\over\partial#2}}$ $\def\parts#1#2#3{{\partial^2#1\over\partial#2\,\partial#3}}$ On the left hand side of your equations, you have the symbol "$\part{\vphantom f}y\bigl(\part f x\bigr)"$. By definition this is the partial derivative of the function $\part fx$ with respect to $y$. So, upon encountering this symbol, you take the function $\part fx$ and then take its partial with respect to $y$. The natural notation of the type on the right hand side of your equations is the notation used in (2) of your post: $$\tag{3} \part{\vphantom f}y\Bigl(\part f x\Bigr)=\parts f y x. $$ I will not surmise why this is the "natural" notation, but will point out that $(3)$ gives the adopted definition for the symbol $\parts f y x$ in any calculus/analysis text, or any other "credible" source, you'll find. I emphasise here that $(3)$ defines the symbol $\parts f y x$; that this sometimes gives an expression that equals $\parts f x y$ is irrelevant. (Of course, for certain functions, what you wrote in (1) would be correct; but its correctness would follow from the result of a theorem, not from the definition of the symbols.) answered Apr 8, 2012 at 17:32 $\endgroup$ 2 1 $\begingroup$ The order is important when the function is not $C^2$. That is, the second derivatives (in relation to any combination of two variables) of $f$ are continuous functions. If the function is $C^2$ then it doesn't matter the order in which the variables appear. This is a widely known result called Schwarz's Theorem, but it seems that there are other names for it. Check out for more in http://en.wikipedia.org/wiki/Symmetry_of_partial_derivatives under the "Clairaut's theorem" subtitle. answered Apr 2, 2012 at 20:03 $\endgroup$ 11 0 $\begingroup$ here i uesd notation $f_{xy}$ and $f_{yx}$ for double derivative.\ There is some theorem on equality of $f_{xy}$ and $f_{yx}$. Theorem : (Young's theorem)\ Let $f$ be a real function defined on non-empty open set $E$ subset of $R^{2}$. If $f_{x}$ and $f_{y}$ exist in some nbhd. of $(x,y)$ and are both differentiable at point $(x,y)$ with respect to $x$ and $y$ then $f_{xy}=f_{yx}$ at point $(x,y)$. Theorem: (Schwartz's theorem) Let $f:E\to R$ be a function such that its partial derivatives $f_{x},f_{y}$ and $f_{xy}$ exist and are continuous in a nbhd. of a point $(x,y)$ then $f_{yx}$ exist such that $f_{xy}=f_{yx}$. Note: One can see that the converse of above theorem need not be true. answered Apr 8, 2012 at 8:51 $\endgroup$ 0 $\begingroup$ I always saw the second : the last operation you made is derivating with respect Y and tha symbol appears in the same order . In the other notation the order is inverse fxy it means the last operation you made is derivating with respect y. But : if both fxy and fyx are defined at a neighborhood of a point and are continuous at the point their value is the same at the point.See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10 answered Apr 8, 2012 at 10:32 $\endgroup$ 1 You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions See similar questions with these tags.
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[Explore Stack Internal](https://stackoverflow.co/internal/?utm_medium=referral&utm_source=math-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover) # [Second order partial derivatives - notation](https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation) [Ask Question](https://math.stackexchange.com/questions/ask) Asked 13 years, 11 months ago Modified [13 years, 11 months ago](https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation?lastactivity "2012-04-08 17:32:24Z") Viewed 5k times 4 \$\\begingroup\$ I have seen both of these used, and people around me seem to disagree, so which one is correct: (first derivative with respect to x, then y): (1) \$\$\\frac{\\partial }{\\partial y}(\\frac{\\partial f}{\\partial x}) = \\frac{\\partial^{2} f}{\\partial x\\partial y}\$\$ (2) \$\$\\frac{\\partial }{\\partial y}(\\frac{\\partial f}{\\partial x}) = \\frac{\\partial^{2} f}{\\partial y\\partial x}\$\$ and why? (reasons, history?) - [calculus](https://math.stackexchange.com/questions/tagged/calculus "show questions tagged 'calculus'") - [multivariable-calculus](https://math.stackexchange.com/questions/tagged/multivariable-calculus "show questions tagged 'multivariable-calculus'") - [notation](https://math.stackexchange.com/questions/tagged/notation "show questions tagged 'notation'") - [derivatives](https://math.stackexchange.com/questions/tagged/derivatives "show questions tagged 'derivatives'") [Share](https://math.stackexchange.com/q/127361 "Short permalink to this question") Cite Follow [edited Apr 4, 2012 at 1:35](https://math.stackexchange.com/posts/127361/revisions "show all edits to this post") asked Apr 2, 2012 at 19:52 [![Thomas's user avatar](https://www.gravatar.com/avatar/e55135a3fbd2d22f1bd86f34cec13d3c?s=64&d=identicon&r=PG)](https://math.stackexchange.com/users/26188/thomas) [Thomas](https://math.stackexchange.com/users/26188/thomas) 44\.9k1313 gold badges7777 silver badges140140 bronze badges \$\\endgroup\$ [Add a comment](https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation "Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.") \\| ## 4 Answers 4 Sorted by: [Reset to default](https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation?answertab=scoredesc#tab-top) 5 \+50 \$\\begingroup\$ \$\\def\\part\#1\#2{{\\partial\#1\\over\\partial\#2}}\$ \$\\def\\parts\#1\#2\#3{{\\partial^2\#1\\over\\partial\#2\\,\\partial\#3}}\$ On the left hand side of your equations, you have the symbol "\$\\part{\\vphantom f}y\\bigl(\\part f x\\bigr)"\$. By definition this is the partial derivative of the function \$\\part fx\$ with respect to \$y\$. So, upon encountering this symbol, you take the function \$\\part fx\$ and then take its partial with respect to \$y\$. The natural notation of the type on the right hand side of your equations is the notation used in (2) of your post: \$\$\\tag{3} \\part{\\vphantom f}y\\Bigl(\\part f x\\Bigr)=\\parts f y x. \$\$ I will not surmise why this is the "natural" notation, but will point out that \$(3)\$ gives the adopted definition for the symbol \$\\parts f y x\$ in any calculus/analysis text, or any other "credible" source, you'll find. I emphasise here that \$(3)\$ defines the symbol \$\\parts f y x\$; that this sometimes gives an expression that equals \$\\parts f x y\$ is irrelevant. (Of course, for certain functions, what you wrote in (1) would be correct; but its correctness would follow from the result of a theorem, not from the definition of the symbols.) [Share](https://math.stackexchange.com/a/129358 "Short permalink to this answer") Cite Follow answered Apr 8, 2012 at 17:32 [![David Mitra's user avatar](https://i.sstatic.net/vJm8f.png?s=64)](https://math.stackexchange.com/users/18986/david-mitra) [David Mitra](https://math.stackexchange.com/users/18986/david-mitra) 76\.8k1111 gold badges149149 silver badges205205 bronze badges \$\\endgroup\$ 2 - \$\\begingroup\$ Ok, so I think that this was the answer that I was looking for. I had seen the other notation in a business calculus book and they were being pretty consistent about it. I had also asked a Ph.D. student (with specialty in analysis) and he said that he used the first (1) notation. Can you think of any reason why someone might want the other notation? (option (1)) \$\\endgroup\$ Thomas – [Thomas](https://math.stackexchange.com/users/26188/thomas "44,901 reputation") 2012-04-08 23:21:11 +00:00 Commented Apr 8, 2012 at 23:21 - 1 \$\\begingroup\$ @Thomas Most, if not all, texts (including [Paul's online notes](http://tutorial.math.lamar.edu/Classes/CalcIII/HighOrderPartialDerivs.aspx), Stewart's Calculus, Apostol's Calculus, etc...) I've seen use (2) as the definition. It does seem natural to me since you can "multiply the left side" to get the right side. I can think of no reason to prefer the other. In any case, the symbol does need a definition, as the two expressions can give different results for not so nice functions. For example, there are differentiable functions with unequal mixed derivatives. \$\\endgroup\$ David Mitra – [David Mitra](https://math.stackexchange.com/users/18986/david-mitra "76,753 reputation") 2012-04-08 23:35:50 +00:00 Commented Apr 8, 2012 at 23:35 [Add a comment](https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \\| 1 \$\\begingroup\$ The order is important when the function is not \$C^2\$. That is, the second derivatives (in relation to any combination of two variables) of \$f\$ are continuous functions. If the function is \$C^2\$ then it doesn't matter the order in which the variables appear. This is a widely known result called Schwarz's Theorem, but it seems that there are other names for it. Check out for more in <http://en.wikipedia.org/wiki/Symmetry_of_partial_derivatives> under the "Clairaut's theorem" subtitle. [Share](https://math.stackexchange.com/a/127369 "Short permalink to this answer") Cite Follow answered Apr 2, 2012 at 20:03 [![Marra's user avatar](https://i.sstatic.net/bjgAJ.jpg?s=64)](https://math.stackexchange.com/users/27071/marra) [Marra](https://math.stackexchange.com/users/27071/marra) 5,0632929 silver badges6868 bronze badges \$\\endgroup\$ 11 - \$\\begingroup\$ please explain all difference between all the cases and how come they are different? \$\\endgroup\$ Victor – [Victor](https://math.stackexchange.com/users/12796/victor "8,616 reputation") 2012-04-02 20:04:28 +00:00 Commented Apr 2, 2012 at 20:04 - 1 \$\\begingroup\$ @Victor Have you taken a multi variable calculus course? If so, it should be clear that they will infact be different for some classes of functions and whatnot\! \$\\endgroup\$ user21436 – user21436 2012-04-02 20:08:00 +00:00 Commented Apr 2, 2012 at 20:08 - \$\\begingroup\$ @KannappanSampath - no, never taken it before. \$\\endgroup\$ Victor – [Victor](https://math.stackexchange.com/users/12796/victor "8,616 reputation") 2012-04-02 20:10:09 +00:00 Commented Apr 2, 2012 at 20:10 - 1 \$\\begingroup\$ @Victor Then I suggest you do it. :P \$\\endgroup\$ user21436 – user21436 2012-04-02 20:10:47 +00:00 Commented Apr 2, 2012 at 20:10 - \$\\begingroup\$ Hi, thanks for the response, but I was asking about which notation is correct. \$\\endgroup\$ Thomas – [Thomas](https://math.stackexchange.com/users/26188/thomas "44,901 reputation") 2012-04-02 20:17:15 +00:00 Commented Apr 2, 2012 at 20:17 \\| [Show 6 more comments](https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation "Expand to show all comments on this post") 0 \$\\begingroup\$ here i uesd notation \$f\_{xy}\$ and \$f\_{yx}\$ for double derivative.\\ There is some theorem on equality of \$f\_{xy}\$ and \$f\_{yx}\$. Theorem : (Young's theorem)\\ Let \$f\$ be a real function defined on non-empty open set \$E\$ subset of \$R^{2}\$. If \$f\_{x}\$ and \$f\_{y}\$ exist in some nbhd. of \$(x,y)\$ and are both differentiable at point \$(x,y)\$ with respect to \$x\$ and \$y\$ then \$f\_{xy}=f\_{yx}\$ at point \$(x,y)\$. Theorem: (Schwartz's theorem) Let \$f:E\\to R\$ be a function such that its partial derivatives \$f\_{x},f\_{y}\$ and \$f\_{xy}\$ exist and are continuous in a nbhd. of a point \$(x,y)\$ then \$f\_{yx}\$ exist such that \$f\_{xy}=f\_{yx}\$. Note: One can see that the converse of above theorem need not be true. [Share](https://math.stackexchange.com/a/129248 "Short permalink to this answer") Cite Follow answered Apr 8, 2012 at 8:51 [![Kns's user avatar](https://www.gravatar.com/avatar/1579adfb80a078396b9688d8aa6b1039?s=64&d=identicon&r=PG)](https://math.stackexchange.com/users/27579/kns) [Kns](https://math.stackexchange.com/users/27579/kns) 3,35322 gold badges3232 silver badges4343 bronze badges \$\\endgroup\$ [Add a comment](https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \\| 0 \$\\begingroup\$ I always saw the second : the last operation you made is derivating with respect Y and tha symbol appears in the same order . In the other notation the order is inverse fxy it means the last operation you made is derivating with respect y. But : if both fxy and fyx are defined at a neighborhood of a point and are continuous at the point their value is the same at the point.See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10 [Share](https://math.stackexchange.com/a/129264 "Short permalink to this answer") Cite Follow [edited Apr 8, 2012 at 11:01](https://math.stackexchange.com/posts/129264/revisions "show all edits to this post") answered Apr 8, 2012 at 10:32 [![alpha.Debi's user avatar](https://www.gravatar.com/avatar/9863491b93c64d5849391d09bcb088f8?s=64&d=identicon&r=PG)](https://math.stackexchange.com/users/20863/alpha-debi) [alpha.Debi](https://math.stackexchange.com/users/20863/alpha-debi) 1,09477 silver badges77 bronze badges \$\\endgroup\$ 1 - \$\\begingroup\$ See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10 \$\\endgroup\$ alpha.Debi – [alpha.Debi](https://math.stackexchange.com/users/20863/alpha-debi "1,094 reputation") 2012-04-08 11:01:03 +00:00 Commented Apr 8, 2012 at 11:01 [Add a comment](https://math.stackexchange.com/questions/127361/second-order-partial-derivatives-notation "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \\| ## You must [log in](https://math.stackexchange.com/users/login?ssrc=question_page&returnurl=https%3A%2F%2Fmath.stackexchange.com%2Fquestions%2F127361) to answer this question. Start asking to get answers Find the answer to your question by asking. 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Readable Markdown	4 \$\\begingroup\$ I have seen both of these used, and people around me seem to disagree, so which one is correct: (first derivative with respect to x, then y): (1) \$\$\\frac{\\partial }{\\partial y}(\\frac{\\partial f}{\\partial x}) = \\frac{\\partial^{2} f}{\\partial x\\partial y}\$\$ (2) \$\$\\frac{\\partial }{\\partial y}(\\frac{\\partial f}{\\partial x}) = \\frac{\\partial^{2} f}{\\partial y\\partial x}\$\$ and why? (reasons, history?) asked Apr 2, 2012 at 19:52 [![Thomas's user avatar](https://www.gravatar.com/avatar/e55135a3fbd2d22f1bd86f34cec13d3c?s=64&d=identicon&r=PG)](https://math.stackexchange.com/users/26188/thomas) \$\\endgroup\$ 5 \+50 \$\\begingroup\$ \$\\def\\part\#1\#2{{\\partial\#1\\over\\partial\#2}}\$ \$\\def\\parts\#1\#2\#3{{\\partial^2\#1\\over\\partial\#2\\,\\partial\#3}}\$ On the left hand side of your equations, you have the symbol "\$\\part{\\vphantom f}y\\bigl(\\part f x\\bigr)"\$. By definition this is the partial derivative of the function \$\\part fx\$ with respect to \$y\$. So, upon encountering this symbol, you take the function \$\\part fx\$ and then take its partial with respect to \$y\$. The natural notation of the type on the right hand side of your equations is the notation used in (2) of your post: \$\$\\tag{3} \\part{\\vphantom f}y\\Bigl(\\part f x\\Bigr)=\\parts f y x. \$\$ I will not surmise why this is the "natural" notation, but will point out that \$(3)\$ gives the adopted definition for the symbol \$\\parts f y x\$ in any calculus/analysis text, or any other "credible" source, you'll find. I emphasise here that \$(3)\$ defines the symbol \$\\parts f y x\$; that this sometimes gives an expression that equals \$\\parts f x y\$ is irrelevant. (Of course, for certain functions, what you wrote in (1) would be correct; but its correctness would follow from the result of a theorem, not from the definition of the symbols.) answered Apr 8, 2012 at 17:32 [![David Mitra's user avatar](https://i.sstatic.net/vJm8f.png?s=64)](https://math.stackexchange.com/users/18986/david-mitra) \$\\endgroup\$ 2 1 \$\\begingroup\$ The order is important when the function is not \$C^2\$. That is, the second derivatives (in relation to any combination of two variables) of \$f\$ are continuous functions. If the function is \$C^2\$ then it doesn't matter the order in which the variables appear. This is a widely known result called Schwarz's Theorem, but it seems that there are other names for it. Check out for more in <http://en.wikipedia.org/wiki/Symmetry_of_partial_derivatives> under the "Clairaut's theorem" subtitle. answered Apr 2, 2012 at 20:03 [![Marra's user avatar](https://i.sstatic.net/bjgAJ.jpg?s=64)](https://math.stackexchange.com/users/27071/marra) \$\\endgroup\$ 11 0 \$\\begingroup\$ here i uesd notation \$f\_{xy}\$ and \$f\_{yx}\$ for double derivative.\\ There is some theorem on equality of \$f\_{xy}\$ and \$f\_{yx}\$. Theorem : (Young's theorem)\\ Let \$f\$ be a real function defined on non-empty open set \$E\$ subset of \$R^{2}\$. If \$f\_{x}\$ and \$f\_{y}\$ exist in some nbhd. of \$(x,y)\$ and are both differentiable at point \$(x,y)\$ with respect to \$x\$ and \$y\$ then \$f\_{xy}=f\_{yx}\$ at point \$(x,y)\$. Theorem: (Schwartz's theorem) Let \$f:E\\to R\$ be a function such that its partial derivatives \$f\_{x},f\_{y}\$ and \$f\_{xy}\$ exist and are continuous in a nbhd. of a point \$(x,y)\$ then \$f\_{yx}\$ exist such that \$f\_{xy}=f\_{yx}\$. Note: One can see that the converse of above theorem need not be true. answered Apr 8, 2012 at 8:51 [![Kns's user avatar](https://www.gravatar.com/avatar/1579adfb80a078396b9688d8aa6b1039?s=64&d=identicon&r=PG)](https://math.stackexchange.com/users/27579/kns) \$\\endgroup\$ 0 \$\\begingroup\$ I always saw the second : the last operation you made is derivating with respect Y and tha symbol appears in the same order . In the other notation the order is inverse fxy it means the last operation you made is derivating with respect y. But : if both fxy and fyx are defined at a neighborhood of a point and are continuous at the point their value is the same at the point.See :Mathematical Analysis , Tom .A. Apostol, Definition, 6-10 answered Apr 8, 2012 at 10:32 [![alpha.Debi's user avatar](https://www.gravatar.com/avatar/9863491b93c64d5849391d09bcb088f8?s=64&d=identicon&r=PG)](https://math.stackexchange.com/users/20863/alpha-debi) \$\\endgroup\$ 1 ## You must [log in](https://math.stackexchange.com/users/login?ssrc=question_page&returnurl=https%3A%2F%2Fmath.stackexchange.com%2Fquestions%2F127361) to answer this question. Start asking to get answers Find the answer to your question by asking. [Ask question](https://math.stackexchange.com/questions/ask) Explore related questions See similar questions with these tags.
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