{"id":3919,"date":"2022-04-05T18:39:12","date_gmt":"2022-04-05T18:39:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3919"},"modified":"2022-10-29T01:20:58","modified_gmt":"2022-10-29T01:20:58","slug":"higher-order-partial-derivatives","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/higher-order-partial-derivatives\/","title":{"raw":"Higher-Order Partial Derivatives","rendered":"Higher-Order Partial Derivatives"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine the higher-order derivatives of a function of two variables.<\/li>\r\n<\/ul>\r\n<\/div>\r\nConsider the function\r\n<p style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y)=2x^{3}-4xy^{2}+5y^{3}-6xy+5x-4y+12}.[\/latex]<\/p>\r\nIts partial derivatives are\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial x}=6x^{2}-4y^{2}-6y+5}[\/latex] and [latex]\\large{\\frac{\\partial f}{\\partial y}=-8xy+15y^{2}-6x-4}.[\/latex]<\/p>\r\nEach of these\u00a0partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions. Just as with derivatives of single-variable functions, we can call these <em data-effect=\"italics\">second-order derivatives, third-order derivatives<\/em>, and so on. In general, they are referred to as <strong><span id=\"b1ffd73f-b9d7-41a1-8546-c25c366b2e90_term168\" data-type=\"term\">higher-order partial derivatives<\/span><\/strong>. There are four second-order partial derivatives for any function (provided they all exist):\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial^{2}f}{\\partial x^{2}}=\\frac{\\partial}{\\partial x}\\left[\\frac{\\partial f}{\\partial x}\\right],\\ \\frac{\\partial^{2}f}{\\partial x\\partial y}=\\frac{\\partial}{\\partial x}\\left[\\frac{\\partial f}{\\partial y}\\right],\\ \\frac{\\partial^{2}f}{\\partial y\\partial x}=\\frac{\\partial}{\\partial y}\\left[\\frac{\\partial f}{\\partial x}\\right],\\ \\frac{\\partial^{2}f}{\\partial y^{2}}=\\frac{\\partial}{\\partial y}\\left[\\frac{\\partial f}{\\partial y}\\right]}.[\/latex]<\/p>\r\nAn alternative notation for each is [latex]f_{xx},\\ f_{yx},\\ f_{xy}[\/latex], and [latex]f_{yy}[\/latex], respectively. Higher-order partial derivatives calculated with respect to different variables, such as [latex]f_{xy}[\/latex] and [latex]f_{yx}[\/latex], are commonly called\u00a0<strong>mixed partial derivatives<\/strong>.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Calculating Second Partial Derivatives<\/h3>\r\nCalculate all four second partial derivates for the function\r\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y)=xe^{-3y}+\\sin{(2x-5y)}}[\/latex].<\/div>\r\n[reveal-answer q=\"fs-id1367983930124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367983930124\"]\r\n<p style=\"text-align: left;\">To calculate [latex]\\partial^{2}f\/\\partial x^{2}[\/latex] and [latex]\\partial^{2}f\/\\partial y\\partial x[\/latex], we first calculate [latex]\\partial f\/\\partial x[\/latex]:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial x}=e^{-3y}+2\\cos{(2x-5y)}}.[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\partial^{2}f\/\\partial x^{2}[\/latex], differentiate [latex]\\partial f\/\\partial x[\/latex] with respect to [latex]x[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{{\\partial}^{2} f}{\\partial {x^{2}}}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\left[\\frac{\\partial f}{\\partial x}\\right]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\left[e^{-3y}+2\\cos{(2x-5y)}\\right]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-4\\sin{(2x-5y)}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\partial^{2}f\/\\partial y\\partial x[\/latex], differentiate [latex]\\partial f\/\\partial x[\/latex] with respect to [latex]y[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial^{2}f}{\\partial y\\partial x}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\left[\\frac{\\partial f}{\\partial x}\\right]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\left[e^{-3y}+2\\cos{(2x-5y)}\\right]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-3e^{-3y}+10\\sin{(2x-5y)}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\partial^{2}f\/\\partial x\\partial y[\/latex] and [latex]\\partial^{2}f\/\\partial y^{2}[\/latex], first calculate [latex]\\partial f\/\\partial y[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}=-3xe^{-3y}-5\\cos{(2x-5y)}.}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\partial^{2}f\/\\partial x\\partial y[\/latex], differentiate [latex]\\partial f\/\\partial y[\/latex] with respect to [latex]x[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial^{2}f}{\\partial x\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\left[\\frac{\\partial f}{\\partial y}\\right]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial x}\\left[-3xe^{-3y}-5\\cos{(2x-5y)}\\right]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-3e^{-3y}+10\\sin{(2x-5y)}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTo calculate [latex]\\partial^{2}f\/\\partial^{2} y[\/latex], differentiate [latex]\\partial f\/\\partial y[\/latex] with respect to [latex]y[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial^{2}f}{\\partial y}} &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\left[\\frac{\\partial f}{\\partial y}\\right]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\partial}{\\partial y}\\left[-3xe^{-3y}-5\\cos{(2x-5y)}\\right]} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {9xe^{-3y}-25\\sin{(2x-5y)}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nCalculate all four second partial derivatives for the function\r\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y)=\\sin{(3x-2y)}+\\cos{(x+4y)}.}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n[reveal-answer q=\"fs-id1567773930324\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1567773930324\"]\r\n<div style=\"text-align: center;\">[latex]\\frac{\\partial^{2}f}{\\partial x^{2}}=-9\\sin{(3x-2y)}-\\cos{(x+4y)}[\/latex]<\/div>\r\n&nbsp;\r\n<div style=\"text-align: center;\">[latex]\\frac{\\partial^{2}f}{\\partial x\\partial y}=6\\sin{(3x-2y)}-4\\cos{(x+4y)}[\/latex]<\/div>\r\n&nbsp;\r\n<div style=\"text-align: center;\">[latex]\\frac{\\partial^{2}f}{\\partial y\\partial x}=6\\sin{(3x-2y)}-4\\cos{(x+4y)}[\/latex]<\/div>\r\n&nbsp;\r\n<div style=\"text-align: center;\">[latex]\\frac{\\partial^{2}f}{\\partial y^{2}}=-4\\sin{(3x-2y)}-16\\cos{(x+4y)}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186155&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=lIhClFtXppQ&amp;video_target=tpm-plugin-k9q1ge88-lIhClFtXppQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.17_transcript.html\"> transcript for \u201cCP 4.17\u201d here (opens in new window).<\/a><\/center>At this point we should notice that, in both the previous example and the \"Try It,\" it was true that [latex]\\partial^{2}f\/\\partial x\\partial y=\\partial^{2}f\/\\partial y\\partial x[\/latex]. Under certain conditions, this is always true. In fact, it is a direct consequence of the following theorem.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Equality of Mixed PArtial Derivatives (Clairaut's Theorem)<\/h3>\r\n\r\n<hr \/>\r\n\r\nSuppose that [latex]f\\,(x,\\ y)[\/latex] is defined on an open disk [latex]D[\/latex] that contains the point [latex](a,\\ b)[\/latex]. If the functions [latex]f_{xy}[\/latex] and [latex]f_{yx}[\/latex] are continuous on [latex]D[\/latex], then [latex]f_{xy}=f_{yx}[\/latex].\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793586299\">Clairaut\u2019s theorem guarantees that as long as mixed second-order derivatives are continuous, the order in which we choose to differentiate the functions (i.e., which variable goes first, then second, and so on) does not matter. It can be extended to higher-order derivatives as well. The proof of Clairaut\u2019s theorem can be found in most advanced calculus books.<\/p>\r\n<p id=\"fs-id1167793586307\">Two other second-order partial derivatives can be calculated for any function [latex]f\\,(x,\\ y)[\/latex]. The partial derivative [latex]f_{xx}[\/latex] is equal to the partial derivative of [latex]f_x[\/latex] with respect to [latex]x[\/latex], and [latex]f_{yy}[\/latex] is equal to the partial derivative of [latex]f_y[\/latex] with respect to [latex]y[\/latex].<\/p>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine the higher-order derivatives of a function of two variables.<\/li>\n<\/ul>\n<\/div>\n<p>Consider the function<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y)=2x^{3}-4xy^{2}+5y^{3}-6xy+5x-4y+12}.[\/latex]<\/p>\n<p>Its partial derivatives are<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial x}=6x^{2}-4y^{2}-6y+5}[\/latex] and [latex]\\large{\\frac{\\partial f}{\\partial y}=-8xy+15y^{2}-6x-4}.[\/latex]<\/p>\n<p>Each of these\u00a0partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions. Just as with derivatives of single-variable functions, we can call these <em data-effect=\"italics\">second-order derivatives, third-order derivatives<\/em>, and so on. In general, they are referred to as <strong><span id=\"b1ffd73f-b9d7-41a1-8546-c25c366b2e90_term168\" data-type=\"term\">higher-order partial derivatives<\/span><\/strong>. There are four second-order partial derivatives for any function (provided they all exist):<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial^{2}f}{\\partial x^{2}}=\\frac{\\partial}{\\partial x}\\left[\\frac{\\partial f}{\\partial x}\\right],\\ \\frac{\\partial^{2}f}{\\partial x\\partial y}=\\frac{\\partial}{\\partial x}\\left[\\frac{\\partial f}{\\partial y}\\right],\\ \\frac{\\partial^{2}f}{\\partial y\\partial x}=\\frac{\\partial}{\\partial y}\\left[\\frac{\\partial f}{\\partial x}\\right],\\ \\frac{\\partial^{2}f}{\\partial y^{2}}=\\frac{\\partial}{\\partial y}\\left[\\frac{\\partial f}{\\partial y}\\right]}.[\/latex]<\/p>\n<p>An alternative notation for each is [latex]f_{xx},\\ f_{yx},\\ f_{xy}[\/latex], and [latex]f_{yy}[\/latex], respectively. Higher-order partial derivatives calculated with respect to different variables, such as [latex]f_{xy}[\/latex] and [latex]f_{yx}[\/latex], are commonly called\u00a0<strong>mixed partial derivatives<\/strong>.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Calculating Second Partial Derivatives<\/h3>\n<p>Calculate all four second partial derivates for the function<\/p>\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y)=xe^{-3y}+\\sin{(2x-5y)}}[\/latex].<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367983930124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367983930124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">To calculate [latex]\\partial^{2}f\/\\partial x^{2}[\/latex] and [latex]\\partial^{2}f\/\\partial y\\partial x[\/latex], we first calculate [latex]\\partial f\/\\partial x[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial x}=e^{-3y}+2\\cos{(2x-5y)}}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\partial^{2}f\/\\partial x^{2}[\/latex], differentiate [latex]\\partial f\/\\partial x[\/latex] with respect to [latex]x[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{{\\partial}^{2} f}{\\partial {x^{2}}}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\left[\\frac{\\partial f}{\\partial x}\\right]} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\partial}{\\partial x}\\left[e^{-3y}+2\\cos{(2x-5y)}\\right]} \\hfill \\\\ \\hfill & =\\hfill & {-4\\sin{(2x-5y)}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\partial^{2}f\/\\partial y\\partial x[\/latex], differentiate [latex]\\partial f\/\\partial x[\/latex] with respect to [latex]y[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial^{2}f}{\\partial y\\partial x}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\left[\\frac{\\partial f}{\\partial x}\\right]} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\partial}{\\partial y}\\left[e^{-3y}+2\\cos{(2x-5y)}\\right]} \\hfill \\\\ \\hfill & =\\hfill & {-3e^{-3y}+10\\sin{(2x-5y)}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\partial^{2}f\/\\partial x\\partial y[\/latex] and [latex]\\partial^{2}f\/\\partial y^{2}[\/latex], first calculate [latex]\\partial f\/\\partial y[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\frac{\\partial f}{\\partial y}=-3xe^{-3y}-5\\cos{(2x-5y)}.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\partial^{2}f\/\\partial x\\partial y[\/latex], differentiate [latex]\\partial f\/\\partial y[\/latex] with respect to [latex]x[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial^{2}f}{\\partial x\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial x}\\left[\\frac{\\partial f}{\\partial y}\\right]} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\partial}{\\partial x}\\left[-3xe^{-3y}-5\\cos{(2x-5y)}\\right]} \\hfill \\\\ \\hfill & =\\hfill & {-3e^{-3y}+10\\sin{(2x-5y)}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>To calculate [latex]\\partial^{2}f\/\\partial^{2} y[\/latex], differentiate [latex]\\partial f\/\\partial y[\/latex] with respect to [latex]y[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\frac{\\partial^{2}f}{\\partial y}} & =\\hfill & {\\frac{\\partial}{\\partial y}\\left[\\frac{\\partial f}{\\partial y}\\right]} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\partial}{\\partial y}\\left[-3xe^{-3y}-5\\cos{(2x-5y)}\\right]} \\hfill \\\\ \\hfill & =\\hfill & {9xe^{-3y}-25\\sin{(2x-5y)}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Calculate all four second partial derivatives for the function<\/p>\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ y)=\\sin{(3x-2y)}+\\cos{(x+4y)}.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1567773930324\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1567773930324\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\frac{\\partial^{2}f}{\\partial x^{2}}=-9\\sin{(3x-2y)}-\\cos{(x+4y)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\partial^{2}f}{\\partial x\\partial y}=6\\sin{(3x-2y)}-4\\cos{(x+4y)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\partial^{2}f}{\\partial y\\partial x}=6\\sin{(3x-2y)}-4\\cos{(x+4y)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\partial^{2}f}{\\partial y^{2}}=-4\\sin{(3x-2y)}-16\\cos{(x+4y)}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186155&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=lIhClFtXppQ&amp;video_target=tpm-plugin-k9q1ge88-lIhClFtXppQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.17_transcript.html\"> transcript for \u201cCP 4.17\u201d here (opens in new window).<\/a><\/div>\n<p>At this point we should notice that, in both the previous example and the &#8220;Try It,&#8221; it was true that [latex]\\partial^{2}f\/\\partial x\\partial y=\\partial^{2}f\/\\partial y\\partial x[\/latex]. Under certain conditions, this is always true. In fact, it is a direct consequence of the following theorem.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Equality of Mixed PArtial Derivatives (Clairaut&#8217;s Theorem)<\/h3>\n<hr \/>\n<p>Suppose that [latex]f\\,(x,\\ y)[\/latex] is defined on an open disk [latex]D[\/latex] that contains the point [latex](a,\\ b)[\/latex]. If the functions [latex]f_{xy}[\/latex] and [latex]f_{yx}[\/latex] are continuous on [latex]D[\/latex], then [latex]f_{xy}=f_{yx}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167793586299\">Clairaut\u2019s theorem guarantees that as long as mixed second-order derivatives are continuous, the order in which we choose to differentiate the functions (i.e., which variable goes first, then second, and so on) does not matter. It can be extended to higher-order derivatives as well. The proof of Clairaut\u2019s theorem can be found in most advanced calculus books.<\/p>\n<p id=\"fs-id1167793586307\">Two other second-order partial derivatives can be calculated for any function [latex]f\\,(x,\\ y)[\/latex]. The partial derivative [latex]f_{xx}[\/latex] is equal to the partial derivative of [latex]f_x[\/latex] with respect to [latex]x[\/latex], and [latex]f_{yy}[\/latex] is equal to the partial derivative of [latex]f_y[\/latex] with respect to [latex]y[\/latex].<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3919\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 4.17. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 4.17\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3919","chapter","type-chapter","status-publish","hentry"],"part":22,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3919","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3919\/revisions"}],"predecessor-version":[{"id":5839,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3919\/revisions\/5839"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/22"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3919\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3919"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3919"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3919"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3919"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}