{"id":1110,"date":"2025-10-15T14:19:00","date_gmt":"2025-10-15T14:19:00","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1110"},"modified":"2025-10-27T20:24:46","modified_gmt":"2025-10-27T20:24:46","slug":"the-chain-rule-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/the-chain-rule-learn-it-2\/","title":{"raw":"The Chain Rule: Learn It 2","rendered":"The Chain Rule: Learn It 2"},"content":{"raw":"<h2 data-type=\"title\">The Generalized Chain Rule<\/h2>\r\n<p class=\"whitespace-normal break-words\">Now that we've seen how to extend the chain rule to functions of two variables, a natural question arises: Can we extend the rule to more than two variables? The answer is yes.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>generalized chain rule<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]w=f(x_1,x_2,\\ldots,x_m)[\/latex] be a differentiable function of [latex]m[\/latex] variables, and let each [latex]x_i=x_i(t_1,t_2,\\ldots,t_n)[\/latex] be a differentiable function of [latex]n[\/latex] variables. Then:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial w}{\\partial t_j}=\\dfrac{\\partial w}{\\partial x_1}\\dfrac{\\partial x_1}{\\partial t_j}+\\dfrac{\\partial w}{\\partial x_2}\\frac{\\partial x_2}{\\partial t_j}+\\cdots+\\dfrac{\\partial w}{\\partial x_m}\\frac{\\partial x_m}{\\partial t_j}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">for any [latex]j\\in{1,2,\\ldots,n}[\/latex].<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">In the next example, we calculate the derivative of a function of three independent variables where each variable depends on two other variables.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate\u00a0[latex]\\partial w\/\\partial u[\/latex] and\u00a0[latex]\\partial w\/\\partial v[\/latex] using the following functions:[latex]\\hspace{9cm}\\begin{align}w&amp;=f(x,y,z)=3x^2-2xy+4z^2 \\\\x&amp;=x(u,v)=e^u\\sin v \\\\y&amp;=y(u,v)=e^y\\cos v \\\\\r\n\r\nz&amp;=z(u,v)=e^u.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[reveal-answer q=\"667124093\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"667124093\"]\r\n\r\nThe formulas for\u00a0[latex]\\partial w\/\\partial u[\/latex] and\u00a0[latex]\\partial w\/\\partial v[\/latex] are\r\n<p style=\"text-align: center;\">[latex]\\frac{\\partial w}{\\partial u}=\\frac{\\partial w}{\\partial x}\\cdot \\frac{\\partial x}{\\partial u}+\\frac{\\partial w}{\\partial y}\\cdot \\frac{\\partial y}{\\partial u}+\\frac{\\partial w}{\\partial z}\\cdot \\frac{\\partial z}{\\partial u}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{\\partial w}{\\partial v}=\\frac{\\partial w}{\\partial x}\\cdot \\frac{\\partial x}{\\partial v}+\\frac{\\partial w}{\\partial y}\\cdot \\frac{\\partial y}{\\partial v}+\\frac{\\partial w}{\\partial z}\\cdot \\frac{\\partial z}{\\partial v}.[\/latex]<\/p>\r\nTherefore, there are nine different partial derivatives that need to be calculated and substituted. We need to calculate each of them:\r\n\r\n[latex]\\hspace{6cm}\\begin{alignat}{2}\r\n\r\n\\frac{\\partial w}{\\partial x} &amp;= 6x -2y &amp;\\quad &amp;\\quad \\frac{\\partial w}{\\partial y} &amp;=-2x &amp;\\quad &amp;\\quad\\frac{\\partial w}{\\partial z} &amp;= 8z \\\\\r\n\r\n\\frac{\\partial x}{\\partial u} &amp;= e^u\\sin v &amp;\\quad &amp;\\quad \\frac{\\partial y}{\\partial u} &amp;=e^u\\cos v &amp;\\quad &amp;\\quad\\frac{\\partial z}{\\partial u} &amp;= e^u \\\\\r\n\r\n\\frac{\\partial x}{\\partial v} &amp;= e^u\\cos v &amp;\\quad &amp;\\quad\\frac{\\partial y}{\\partial v} &amp;=-e^u\\sin v &amp;\\quad &amp;\\quad \\frac{\\partial z}{\\partial v} &amp;= 0.\r\n\r\n\\end{alignat}[\/latex]\r\n\r\nNow, we substitute each of them into the first formula to calculate\u00a0[latex]\\partial w\/\\partial u[\/latex]:\r\n\r\n[latex]\\hspace{5cm} \\begin{align}\r\n\r\n\\frac{\\partial w}{\\partial u}&amp;=\\frac{\\partial w}{\\partial x}\\cdot \\frac{\\partial x}{\\partial u}+\\frac{\\partial w}{\\partial y}\\cdot \\frac{\\partial y}{\\partial u}+\\frac{\\partial w}{\\partial z}\\cdot \\frac{\\partial z}{\\partial u} \\\\\r\n\r\n&amp;=(6x-2y)e^u\\sin v-2xe^u\\cos v+8ze^u,\r\n\r\n\\end{align}[\/latex]\r\n\r\nthen substitute [latex]x(u,v)=e^u\\sin v,y(u,v)=e^u\\cos v[\/latex], and [latex]z(u,v)=e^u[\/latex], into this equation:\r\n\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\n\\frac{\\partial w}{\\partial u}&amp;=(6x-2y)e^u\\sin v-2xe^u\\cos v+8ze^u \\\\\r\n\r\n&amp;=(6e^u\\sin v-2e^u\\cos v)e^u\\sin v-2(e^u\\sin v)e^u\\cos v+8e^{2u} \\\\\r\n\r\n&amp;=6e^{2u}\\sin^2{v}-4e^{2u}\\sin v\\cos v+8e^{2u} \\\\\r\n\r\n&amp;=2e^{2u}(3\\sin^2{v}-2\\sin v\\cos v+4).\r\n\r\n\\end{align}[\/latex]\r\n\r\nNext, we calculate\u00a0[latex]\\partial w\/\\partial v[\/latex]:\r\n\r\n[latex]\\hspace{5cm} \\begin{align}\r\n\r\n\\frac{\\partial w}{\\partial v}&amp;=\\frac{\\partial w}{\\partial x}\\cdot \\frac{\\partial x}{\\partial v}+\\frac{\\partial w}{\\partial y}\\cdot \\frac{\\partial y}{\\partial v}+\\frac{\\partial w}{\\partial z}\\cdot \\frac{\\partial z}{\\partial v} \\\\\r\n\r\n&amp;=(6x-2y)e^u\\cos v-2x(-e^u\\sin v)+8z(0),\r\n\r\n\\end{align}[\/latex]\r\n\r\nthen we substitute [latex]x(u,v)=e^u\\sin v,y(u,v)=e^u\\cos v,[\/latex] and\u00a0[latex]z(u,v)=e^u[\/latex] into this equation:\r\n\r\n[latex]\r\n\r\n\\hspace{5cm}\\begin{align}\r\n\r\n\\frac{\\partial w}{\\partial v}&amp;=(6x-2y)e^u\\cos v-2x(-e^u\\sin v) \\\\\r\n\r\n&amp;=(6e^u\\sin v-2e^u\\cos v)e^u\\cos v+2(e^u\\sin v)(e^u\\sin) \\\\\r\n\r\n&amp;=2e^{2u}\\sin^2{v}+6e^{2u}\\sin v\\cos v-2e^{2u}\\cos^2{v} \\\\\r\n\r\n&amp;=2e^{2u}(\\sin^2{v}+\\sin v\\cos v-\\cos^2{v}).\r\n\r\n\\end{align}\r\n\r\n[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Create a tree diagram for the case when\r\n<p style=\"text-align: center;\">[latex]\\large{w=f(x,y,z),x=x(t,u,v),y=y(t,u,v),z=z(t,u,v)}[\/latex]<\/p>\r\nand write out the formulas for the three partial derivatives of\u00a0[latex]w[\/latex].\r\n\r\n[reveal-answer q=\"348851212\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348851212\"]\r\n\r\nStarting from the left, the function [latex]f[\/latex]\u00a0has three independent variables: [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex]. Therefore, three branches must be emanating from the first node. Each of these three branches also has three branches, for each of the variables\u00a0[latex]t[\/latex], [latex]u[\/latex], and [latex]v[\/latex].\r\n\r\n[caption id=\"attachment_1248\" align=\"aligncenter\" width=\"587\"]<img class=\"size-full wp-image-1248\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22221925\/4-5-3.jpeg\" alt=\"A diagram that starts with w = f(x, y, z). Along the first branch, it is written \u2202w\/\u2202x, then x = x(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then \u2202w\/\u2202x \u2202x\/\u2202t; the second subbranch says u and then \u2202w\/\u2202x \u2202x\/\u2202u; and the third subbranch says v and then \u2202w\/\u2202x \u2202x\/\u2202v. Along the second branch, it is written \u2202w\/\u2202y, then y = y(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then \u2202w\/\u2202y \u2202y\/\u2202t; the second subbranch says u and then \u2202w\/\u2202y \u2202y\/\u2202u; and the third subbranch says v and then \u2202w\/\u2202y \u2202y\/\u2202v. Along the third branch, it is written \u2202w\/\u2202z, then z = z(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then \u2202w\/\u2202z \u2202z\/\u2202t; the second subbranch says u and then \u2202w\/\u2202z \u2202z\/\u2202u; and the third subbranch says v and then \u2202w\/\u2202z \u2202z\/\u2202v.\" width=\"587\" height=\"551\" \/> Figure 3. Tree diagram for a function of three variables, each of which is a function of three independent variables.[\/caption]\r\n\r\nThe three formulas are\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial w}{\\partial t}=\\frac{\\partial w}{\\partial x}\\frac{\\partial x}{\\partial t}+\\frac{\\partial w}{\\partial y}\\frac{\\partial y}{\\partial t}+\\frac{\\partial w}{\\partial z}\\frac{\\partial z}{\\partial t}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial w}{\\partial u}=\\frac{\\partial w}{\\partial x}\\frac{\\partial x}{\\partial u}+\\frac{\\partial w}{\\partial y}\\frac{\\partial y}{\\partial u}+\\frac{\\partial w}{\\partial z}\\frac{\\partial z}{\\partial u}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial w}{\\partial v}=\\frac{\\partial w}{\\partial x}\\frac{\\partial x}{\\partial v}+\\frac{\\partial w}{\\partial y}\\frac{\\partial y}{\\partial v}+\\frac{\\partial w}{\\partial z}\\frac{\\partial z}{\\partial v}}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<h2 data-type=\"title\">The Generalized Chain Rule<\/h2>\n<p class=\"whitespace-normal break-words\">Now that we&#8217;ve seen how to extend the chain rule to functions of two variables, a natural question arises: Can we extend the rule to more than two variables? The answer is yes.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>generalized chain rule<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]w=f(x_1,x_2,\\ldots,x_m)[\/latex] be a differentiable function of [latex]m[\/latex] variables, and let each [latex]x_i=x_i(t_1,t_2,\\ldots,t_n)[\/latex] be a differentiable function of [latex]n[\/latex] variables. Then:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial w}{\\partial t_j}=\\dfrac{\\partial w}{\\partial x_1}\\dfrac{\\partial x_1}{\\partial t_j}+\\dfrac{\\partial w}{\\partial x_2}\\frac{\\partial x_2}{\\partial t_j}+\\cdots+\\dfrac{\\partial w}{\\partial x_m}\\frac{\\partial x_m}{\\partial t_j}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">for any [latex]j\\in{1,2,\\ldots,n}[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">In the next example, we calculate the derivative of a function of three independent variables where each variable depends on two other variables.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate\u00a0[latex]\\partial w\/\\partial u[\/latex] and\u00a0[latex]\\partial w\/\\partial v[\/latex] using the following functions:[latex]\\hspace{9cm}\\begin{align}w&=f(x,y,z)=3x^2-2xy+4z^2 \\\\x&=x(u,v)=e^u\\sin v \\\\y&=y(u,v)=e^y\\cos v \\\\    z&=z(u,v)=e^u.    \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q667124093\">Show Solution<\/button><\/p>\n<div id=\"q667124093\" class=\"hidden-answer\" style=\"display: none\">\n<p>The formulas for\u00a0[latex]\\partial w\/\\partial u[\/latex] and\u00a0[latex]\\partial w\/\\partial v[\/latex] are<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\partial w}{\\partial u}=\\frac{\\partial w}{\\partial x}\\cdot \\frac{\\partial x}{\\partial u}+\\frac{\\partial w}{\\partial y}\\cdot \\frac{\\partial y}{\\partial u}+\\frac{\\partial w}{\\partial z}\\cdot \\frac{\\partial z}{\\partial u}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\partial w}{\\partial v}=\\frac{\\partial w}{\\partial x}\\cdot \\frac{\\partial x}{\\partial v}+\\frac{\\partial w}{\\partial y}\\cdot \\frac{\\partial y}{\\partial v}+\\frac{\\partial w}{\\partial z}\\cdot \\frac{\\partial z}{\\partial v}.[\/latex]<\/p>\n<p>Therefore, there are nine different partial derivatives that need to be calculated and substituted. We need to calculate each of them:<\/p>\n<p>[latex]\\hspace{6cm}\\begin{alignat}{2}    \\frac{\\partial w}{\\partial x} &= 6x -2y &\\quad &\\quad \\frac{\\partial w}{\\partial y} &=-2x &\\quad &\\quad\\frac{\\partial w}{\\partial z} &= 8z \\\\    \\frac{\\partial x}{\\partial u} &= e^u\\sin v &\\quad &\\quad \\frac{\\partial y}{\\partial u} &=e^u\\cos v &\\quad &\\quad\\frac{\\partial z}{\\partial u} &= e^u \\\\    \\frac{\\partial x}{\\partial v} &= e^u\\cos v &\\quad &\\quad\\frac{\\partial y}{\\partial v} &=-e^u\\sin v &\\quad &\\quad \\frac{\\partial z}{\\partial v} &= 0.    \\end{alignat}[\/latex]<\/p>\n<p>Now, we substitute each of them into the first formula to calculate\u00a0[latex]\\partial w\/\\partial u[\/latex]:<\/p>\n<p>[latex]\\hspace{5cm} \\begin{align}    \\frac{\\partial w}{\\partial u}&=\\frac{\\partial w}{\\partial x}\\cdot \\frac{\\partial x}{\\partial u}+\\frac{\\partial w}{\\partial y}\\cdot \\frac{\\partial y}{\\partial u}+\\frac{\\partial w}{\\partial z}\\cdot \\frac{\\partial z}{\\partial u} \\\\    &=(6x-2y)e^u\\sin v-2xe^u\\cos v+8ze^u,    \\end{align}[\/latex]<\/p>\n<p>then substitute [latex]x(u,v)=e^u\\sin v,y(u,v)=e^u\\cos v[\/latex], and [latex]z(u,v)=e^u[\/latex], into this equation:<\/p>\n<p>[latex]\\hspace{5cm}\\begin{align}    \\frac{\\partial w}{\\partial u}&=(6x-2y)e^u\\sin v-2xe^u\\cos v+8ze^u \\\\    &=(6e^u\\sin v-2e^u\\cos v)e^u\\sin v-2(e^u\\sin v)e^u\\cos v+8e^{2u} \\\\    &=6e^{2u}\\sin^2{v}-4e^{2u}\\sin v\\cos v+8e^{2u} \\\\    &=2e^{2u}(3\\sin^2{v}-2\\sin v\\cos v+4).    \\end{align}[\/latex]<\/p>\n<p>Next, we calculate\u00a0[latex]\\partial w\/\\partial v[\/latex]:<\/p>\n<p>[latex]\\hspace{5cm} \\begin{align}    \\frac{\\partial w}{\\partial v}&=\\frac{\\partial w}{\\partial x}\\cdot \\frac{\\partial x}{\\partial v}+\\frac{\\partial w}{\\partial y}\\cdot \\frac{\\partial y}{\\partial v}+\\frac{\\partial w}{\\partial z}\\cdot \\frac{\\partial z}{\\partial v} \\\\    &=(6x-2y)e^u\\cos v-2x(-e^u\\sin v)+8z(0),    \\end{align}[\/latex]<\/p>\n<p>then we substitute [latex]x(u,v)=e^u\\sin v,y(u,v)=e^u\\cos v,[\/latex] and\u00a0[latex]z(u,v)=e^u[\/latex] into this equation:<\/p>\n<p>[latex]\\hspace{5cm}\\begin{align}    \\frac{\\partial w}{\\partial v}&=(6x-2y)e^u\\cos v-2x(-e^u\\sin v) \\\\    &=(6e^u\\sin v-2e^u\\cos v)e^u\\cos v+2(e^u\\sin v)(e^u\\sin) \\\\    &=2e^{2u}\\sin^2{v}+6e^{2u}\\sin v\\cos v-2e^{2u}\\cos^2{v} \\\\    &=2e^{2u}(\\sin^2{v}+\\sin v\\cos v-\\cos^2{v}).    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Create a tree diagram for the case when<\/p>\n<p style=\"text-align: center;\">[latex]\\large{w=f(x,y,z),x=x(t,u,v),y=y(t,u,v),z=z(t,u,v)}[\/latex]<\/p>\n<p>and write out the formulas for the three partial derivatives of\u00a0[latex]w[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q348851212\">Show Solution<\/button><\/p>\n<div id=\"q348851212\" class=\"hidden-answer\" style=\"display: none\">\n<p>Starting from the left, the function [latex]f[\/latex]\u00a0has three independent variables: [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex]. Therefore, three branches must be emanating from the first node. Each of these three branches also has three branches, for each of the variables\u00a0[latex]t[\/latex], [latex]u[\/latex], and [latex]v[\/latex].<\/p>\n<figure id=\"attachment_1248\" aria-describedby=\"caption-attachment-1248\" style=\"width: 587px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1248\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22221925\/4-5-3.jpeg\" alt=\"A diagram that starts with w = f(x, y, z). Along the first branch, it is written \u2202w\/\u2202x, then x = x(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then \u2202w\/\u2202x \u2202x\/\u2202t; the second subbranch says u and then \u2202w\/\u2202x \u2202x\/\u2202u; and the third subbranch says v and then \u2202w\/\u2202x \u2202x\/\u2202v. Along the second branch, it is written \u2202w\/\u2202y, then y = y(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then \u2202w\/\u2202y \u2202y\/\u2202t; the second subbranch says u and then \u2202w\/\u2202y \u2202y\/\u2202u; and the third subbranch says v and then \u2202w\/\u2202y \u2202y\/\u2202v. Along the third branch, it is written \u2202w\/\u2202z, then z = z(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then \u2202w\/\u2202z \u2202z\/\u2202t; the second subbranch says u and then \u2202w\/\u2202z \u2202z\/\u2202u; and the third subbranch says v and then \u2202w\/\u2202z \u2202z\/\u2202v.\" width=\"587\" height=\"551\" \/><figcaption id=\"caption-attachment-1248\" class=\"wp-caption-text\">Figure 3. Tree diagram for a function of three variables, each of which is a function of three independent variables.<\/figcaption><\/figure>\n<p>The three formulas are<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial w}{\\partial t}=\\frac{\\partial w}{\\partial x}\\frac{\\partial x}{\\partial t}+\\frac{\\partial w}{\\partial y}\\frac{\\partial y}{\\partial t}+\\frac{\\partial w}{\\partial z}\\frac{\\partial z}{\\partial t}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial w}{\\partial u}=\\frac{\\partial w}{\\partial x}\\frac{\\partial x}{\\partial u}+\\frac{\\partial w}{\\partial y}\\frac{\\partial y}{\\partial u}+\\frac{\\partial w}{\\partial z}\\frac{\\partial z}{\\partial u}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{\\partial w}{\\partial v}=\\frac{\\partial w}{\\partial x}\\frac{\\partial x}{\\partial v}+\\frac{\\partial w}{\\partial y}\\frac{\\partial y}{\\partial v}+\\frac{\\partial w}{\\partial z}\\frac{\\partial z}{\\partial v}}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1110","chapter","type-chapter","status-publish","hentry"],"part":428,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1110","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1110\/revisions"}],"predecessor-version":[{"id":1786,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1110\/revisions\/1786"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/428"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1110\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1110"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1110"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1110"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1110"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}