{"id":618,"date":"2025-10-09T15:25:23","date_gmt":"2025-10-09T15:25:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=618"},"modified":"2025-10-29T14:17:14","modified_gmt":"2025-10-29T14:17:14","slug":"the-chain-rule-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/the-chain-rule-fresh-take\/","title":{"raw":"The Chain Rule: Fresh Take","rendered":"The Chain Rule: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the chain rule when you have one independent variable<\/li>\r\n \t<li>Apply the chain rule with multiple changing variables<\/li>\r\n \t<li>Find derivatives implicitly when working with functions of multiple variables<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Chain Rules for One or Two Independent Variables<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The chain rule extends to multivariable functions, but now we need different versions depending on how many independent variables we have.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Chain Rule for One Independent Variable<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]z = f(x, y)[\/latex] where [latex]x = g(t)[\/latex] and [latex]y = h(t)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{dz}{dt}=\\dfrac{\\partial z}{\\partial x}\\cdot\\dfrac{dx}{dt}+\\dfrac{\\partial z}{\\partial y}\\cdot\\dfrac{dy}{dt}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Notice: We have two terms because [latex]z[\/latex] depends on two intermediate variables ([latex]x[\/latex] and [latex]y[\/latex]). Each term represents one path from [latex]z[\/latex] to [latex]t[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Chain Rule for Two Independent Variables<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]z = f(x, y)[\/latex] where [latex]x = g(u, v)[\/latex] and [latex]y = h(u, v)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial z}{\\partial u}=\\dfrac{\\partial z}{\\partial x}\\dfrac{\\partial x}{\\partial u}+\\dfrac{\\partial z}{\\partial y}\\dfrac{\\partial y}{\\partial u}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial z}{\\partial v}=\\dfrac{\\partial z}{\\partial x}\\dfrac{\\partial x}{\\partial v}+\\dfrac{\\partial z}{\\partial y}\\dfrac{\\partial y}{\\partial v}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Difference<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">One independent variable: Use ordinary derivative [latex]\\frac{dz}{dt}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Two independent variables: Use partial derivatives [latex]\\frac{\\partial z}{\\partial u}[\/latex] and [latex]\\frac{\\partial z}{\\partial v}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Tree Diagrams<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Draw branches showing all paths from [latex]z[\/latex] to the final variable. Each complete path gives one term. Add all terms together.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Calculate all needed partial derivatives. Apply the chain rule formula. Substitute the intermediate variable definitions. Simplify the result.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Calculate [latex]dz\/dt[\/latex] given the following functions. Express the final answer in terms of [latex]t[\/latex].\r\n<p style=\"text-align: center;\">[latex]z=f(x, y)=x^{2}-3xy+2y^{2}, x=x(t)=3\\sin 2t, y=y(t)=4\\cos 2t[\/latex]<\/p>\r\n[reveal-answer q=\"142579180\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"142579180\"]\r\n\r\n[latex]\\hspace{5cm} \\begin{align}\r\n\r\n&amp;=(2x-3y)(6\\cos 2t)+(-3x+y4)(-8\\sin 2t) \\\\\r\n\r\n&amp;=-92\\sin 2t\\cos 2t - 72(\\cos^2{2t}-\\sin^2{2t} \\\\\r\n\r\n&amp;=-46\\sin 4t - 72\\cos 4t.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Calculate\u00a0[latex]\\partial x\/\\partial u[\/latex] and\u00a0[latex]\\partial z\/\\partial v[\/latex] given the following functions:\r\n<p style=\"text-align: center;\">[latex]z=f(x,y)=\\frac{2x-y}{x+3y},x(u,v)=e^{2u}\\cos 3v, y(u,v)=e^{2u}\\sin 3v[\/latex]<\/p>\r\n[reveal-answer q=\"825766032\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"825766032\"]\r\n<p style=\"text-align: center;\">[latex]\\frac{\\partial z}{\\partial u}=0, \\frac{\\partial z}{\\partial v}=\\frac{-21}{(3\\sin 3v + \\cos 3v)^2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186160&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=IBYLL6KOi80&amp;video_target=tpm-plugin-po9x6dci-IBYLL6KOi80\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.24_transcript.html\">transcript for \u201cCP 4.24\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">The Generalized Chain Rule<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The chain rule extends to any number of variables. The pattern stays consistent: add one term for each intermediate variable.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The General Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Let [latex]w = f(x_1, x_2, \\ldots, x_m)[\/latex] depend on [latex]m[\/latex] variables, where each [latex]x_i[\/latex] depends on [latex]n[\/latex] variables [latex]t_1, t_2, \\ldots, t_n[\/latex]:<\/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}\\dfrac{\\partial x_2}{\\partial t_j}+\\cdots+\\dfrac{\\partial w}{\\partial x_m}\\dfrac{\\partial x_m}{\\partial t_j}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The number of terms equals the number of intermediate variables. If [latex]w[\/latex] depends on three intermediate variables ([latex]x[\/latex], [latex]y[\/latex], [latex]z[\/latex]), you get three terms.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Tree Diagrams Help<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Draw branches from [latex]w[\/latex] to each intermediate variable ([latex]x[\/latex], [latex]y[\/latex], [latex]z[\/latex]). From each intermediate variable, draw branches to each final variable ([latex]u[\/latex], [latex]v[\/latex]). Each complete path gives one term in the formula.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Identify all intermediate variables. Calculate all needed partial derivatives. Apply the chain rule formula\u2014one term per intermediate variable. Substitute and simplify.<\/p>\r\n\r\n<\/div>\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] given the following functions:[latex]\\hspace{8cm} \\begin{align}w&amp;=f(x,y,z)=\\frac{x+2y-4z}{2x-y+3z} \\\\\r\n\r\nx&amp;=x(u,v) = e^{2u}\\cos 3v \\\\\r\n\r\ny&amp;=y(u,v) = e^{2u}\\sin 3v \\\\\r\n\r\nz&amp;=z(u,v) = e^{2u}.\r\n\r\n\\end{align}[\/latex]\r\n\r\n[reveal-answer q=\"135764289\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"135764289\"]\r\n\r\n[latex]\\hspace{8cm} \\begin{align}\r\n\r\n\\frac{\\partial w}{\\partial u}&amp;=0 \\\\\r\n\r\n\\frac{\\partial w}{\\partial v}&amp;=\\frac{15-33\\sin 3v+6\\cos 3v}{(3+2 \\cos 3v-\\sin 3v)^2}\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186161&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=roL0IpV1FBU&amp;video_target=tpm-plugin-d4ysvvp8-roL0IpV1FBU\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.25_transcript.html\">transcript for \"CP 4.25\u201d here (opens in new window).<\/a><\/center><\/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),x=x(t,u,v),y=y(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=\"567910368\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567910368\"]\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}}[\/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}}[\/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}}.[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_1253\" align=\"aligncenter\" width=\"437\"]<img class=\"size-full wp-image-1253\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22222504\/4-5-tryitans1.jpeg\" alt=\"A diagram that starts with w = f(x, y). 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.\" width=\"437\" height=\"328\" \/> Figure 4.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2>Implicit Differentiation<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Partial derivatives give us a shortcut for implicit differentiation. Instead of differentiating the entire equation and solving, just calculate two partial derivatives and plug them into a formula.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>For Two Variables<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]f(x, y) = 0[\/latex] defines [latex]y[\/latex] implicitly as a function of [latex]x[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{dy}{dx}=-\\dfrac{\\partial f\/\\partial x}{\\partial f\/\\partial y}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">(provided [latex]f_y \\neq 0[\/latex])<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>For Three Variables<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]f(x, y, z) = 0[\/latex] defines [latex]z[\/latex] implicitly as a function of [latex]x[\/latex] and [latex]y[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial z}{\\partial x}=-\\dfrac{\\partial f\/\\partial x}{\\partial f\/\\partial z}[\/latex] and [latex]\\dfrac{\\partial z}{\\partial y}=-\\dfrac{\\partial f\/\\partial y}{\\partial f\/\\partial z}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">(provided [latex]f_z \\neq 0[\/latex])<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The formula comes from the chain rule. When you differentiate [latex]f(x, y) = 0[\/latex] with respect to [latex]x[\/latex], you get [latex]f_x + f_y \\frac{dy}{dx} = 0[\/latex]. Solving for [latex]\\frac{dy}{dx}[\/latex] gives the formula.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Define [latex]f[\/latex] using the left side of the equation. Calculate the needed partial derivatives. Plug into the formula. Simplify.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find\u00a0[latex]dy\/dx[\/latex] if\u00a0[latex]y[\/latex] is defined implicitly as a function of\u00a0[latex]x[\/latex] by the equation\u00a0[latex]x^{2}+xy-y^{2}+7x-3y-26=0[\/latex]. What is the equation of the tangent line to the graph of this curve at point\u00a0[latex](3, -2)[\/latex]?[reveal-answer q=\"995467213\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"995467213\"]\r\n<p style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\left.\\frac{2x+y+7}{2y-x+3}\\right|_{(3,-2)}=\\frac{2(3)+(-2)+7}{2(-2)-(3)+3}=-\\frac{11}4[\/latex]<\/p>\r\nEquation of the tangent line:\u00a0[latex]y=-\\frac{11}4x+\\frac{25}4[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186162&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=YSpLF3QfzK0&amp;video_target=tpm-plugin-wh05pp33-YSpLF3QfzK0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.27_transcript.html\">transcript for \u201cCP 4.27\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the chain rule when you have one independent variable<\/li>\n<li>Apply the chain rule with multiple changing variables<\/li>\n<li>Find derivatives implicitly when working with functions of multiple variables<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Chain Rules for One or Two Independent Variables<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The chain rule extends to multivariable functions, but now we need different versions depending on how many independent variables we have.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Chain Rule for One Independent Variable<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]z = f(x, y)[\/latex] where [latex]x = g(t)[\/latex] and [latex]y = h(t)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{dz}{dt}=\\dfrac{\\partial z}{\\partial x}\\cdot\\dfrac{dx}{dt}+\\dfrac{\\partial z}{\\partial y}\\cdot\\dfrac{dy}{dt}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Notice: We have two terms because [latex]z[\/latex] depends on two intermediate variables ([latex]x[\/latex] and [latex]y[\/latex]). Each term represents one path from [latex]z[\/latex] to [latex]t[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Chain Rule for Two Independent Variables<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]z = f(x, y)[\/latex] where [latex]x = g(u, v)[\/latex] and [latex]y = h(u, v)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial z}{\\partial u}=\\dfrac{\\partial z}{\\partial x}\\dfrac{\\partial x}{\\partial u}+\\dfrac{\\partial z}{\\partial y}\\dfrac{\\partial y}{\\partial u}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial z}{\\partial v}=\\dfrac{\\partial z}{\\partial x}\\dfrac{\\partial x}{\\partial v}+\\dfrac{\\partial z}{\\partial y}\\dfrac{\\partial y}{\\partial v}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Difference<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">One independent variable: Use ordinary derivative [latex]\\frac{dz}{dt}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Two independent variables: Use partial derivatives [latex]\\frac{\\partial z}{\\partial u}[\/latex] and [latex]\\frac{\\partial z}{\\partial v}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Tree Diagrams<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Draw branches showing all paths from [latex]z[\/latex] to the final variable. Each complete path gives one term. Add all terms together.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Calculate all needed partial derivatives. Apply the chain rule formula. Substitute the intermediate variable definitions. Simplify the result.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate [latex]dz\/dt[\/latex] given the following functions. Express the final answer in terms of [latex]t[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]z=f(x, y)=x^{2}-3xy+2y^{2}, x=x(t)=3\\sin 2t, y=y(t)=4\\cos 2t[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q142579180\">Show Solution<\/button><\/p>\n<div id=\"q142579180\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\hspace{5cm} \\begin{align}    &=(2x-3y)(6\\cos 2t)+(-3x+y4)(-8\\sin 2t) \\\\    &=-92\\sin 2t\\cos 2t - 72(\\cos^2{2t}-\\sin^2{2t} \\\\    &=-46\\sin 4t - 72\\cos 4t.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate\u00a0[latex]\\partial x\/\\partial u[\/latex] and\u00a0[latex]\\partial z\/\\partial v[\/latex] given the following functions:<\/p>\n<p style=\"text-align: center;\">[latex]z=f(x,y)=\\frac{2x-y}{x+3y},x(u,v)=e^{2u}\\cos 3v, y(u,v)=e^{2u}\\sin 3v[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q825766032\">Show Solution<\/button><\/p>\n<div id=\"q825766032\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\frac{\\partial z}{\\partial u}=0, \\frac{\\partial z}{\\partial v}=\\frac{-21}{(3\\sin 3v + \\cos 3v)^2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186160&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=IBYLL6KOi80&amp;video_target=tpm-plugin-po9x6dci-IBYLL6KOi80\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP4.24_transcript.html\">transcript for \u201cCP 4.24\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">The Generalized Chain Rule<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The chain rule extends to any number of variables. The pattern stays consistent: add one term for each intermediate variable.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The General Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Let [latex]w = f(x_1, x_2, \\ldots, x_m)[\/latex] depend on [latex]m[\/latex] variables, where each [latex]x_i[\/latex] depends on [latex]n[\/latex] variables [latex]t_1, t_2, \\ldots, t_n[\/latex]:<\/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}\\dfrac{\\partial x_2}{\\partial t_j}+\\cdots+\\dfrac{\\partial w}{\\partial x_m}\\dfrac{\\partial x_m}{\\partial t_j}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The number of terms equals the number of intermediate variables. If [latex]w[\/latex] depends on three intermediate variables ([latex]x[\/latex], [latex]y[\/latex], [latex]z[\/latex]), you get three terms.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Tree Diagrams Help<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Draw branches from [latex]w[\/latex] to each intermediate variable ([latex]x[\/latex], [latex]y[\/latex], [latex]z[\/latex]). From each intermediate variable, draw branches to each final variable ([latex]u[\/latex], [latex]v[\/latex]). Each complete path gives one term in the formula.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Identify all intermediate variables. Calculate all needed partial derivatives. Apply the chain rule formula\u2014one term per intermediate variable. Substitute and simplify.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Calculate\u00a0[latex]\\partial w\/\\partial u[\/latex] and\u00a0[latex]\\partial w\/\\partial v[\/latex] given the following functions:[latex]\\hspace{8cm} \\begin{align}w&=f(x,y,z)=\\frac{x+2y-4z}{2x-y+3z} \\\\    x&=x(u,v) = e^{2u}\\cos 3v \\\\    y&=y(u,v) = e^{2u}\\sin 3v \\\\    z&=z(u,v) = e^{2u}.    \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q135764289\">Show Solution<\/button><\/p>\n<div id=\"q135764289\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\hspace{8cm} \\begin{align}    \\frac{\\partial w}{\\partial u}&=0 \\\\    \\frac{\\partial w}{\\partial v}&=\\frac{15-33\\sin 3v+6\\cos 3v}{(3+2 \\cos 3v-\\sin 3v)^2}    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186161&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=roL0IpV1FBU&amp;video_target=tpm-plugin-d4ysvvp8-roL0IpV1FBU\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP4.25_transcript.html\">transcript for &#8220;CP 4.25\u201d here (opens in new window).<\/a><\/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),x=x(t,u,v),y=y(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=\"q567910368\">Show Solution<\/button><\/p>\n<div id=\"q567910368\" class=\"hidden-answer\" style=\"display: none\">\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}}[\/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}}[\/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}}.[\/latex]<\/p>\n<figure id=\"attachment_1253\" aria-describedby=\"caption-attachment-1253\" style=\"width: 437px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1253\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22222504\/4-5-tryitans1.jpeg\" alt=\"A diagram that starts with w = f(x, y). 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.\" width=\"437\" height=\"328\" \/><figcaption id=\"caption-attachment-1253\" class=\"wp-caption-text\">Figure 4.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<h2>Implicit Differentiation<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Partial derivatives give us a shortcut for implicit differentiation. Instead of differentiating the entire equation and solving, just calculate two partial derivatives and plug them into a formula.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>For Two Variables<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]f(x, y) = 0[\/latex] defines [latex]y[\/latex] implicitly as a function of [latex]x[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{dy}{dx}=-\\dfrac{\\partial f\/\\partial x}{\\partial f\/\\partial y}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">(provided [latex]f_y \\neq 0[\/latex])<\/p>\n<p class=\"whitespace-normal break-words\"><strong>For Three Variables<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]f(x, y, z) = 0[\/latex] defines [latex]z[\/latex] implicitly as a function of [latex]x[\/latex] and [latex]y[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\dfrac{\\partial z}{\\partial x}=-\\dfrac{\\partial f\/\\partial x}{\\partial f\/\\partial z}[\/latex] and [latex]\\dfrac{\\partial z}{\\partial y}=-\\dfrac{\\partial f\/\\partial y}{\\partial f\/\\partial z}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">(provided [latex]f_z \\neq 0[\/latex])<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The formula comes from the chain rule. When you differentiate [latex]f(x, y) = 0[\/latex] with respect to [latex]x[\/latex], you get [latex]f_x + f_y \\frac{dy}{dx} = 0[\/latex]. Solving for [latex]\\frac{dy}{dx}[\/latex] gives the formula.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Define [latex]f[\/latex] using the left side of the equation. Calculate the needed partial derivatives. Plug into the formula. Simplify.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find\u00a0[latex]dy\/dx[\/latex] if\u00a0[latex]y[\/latex] is defined implicitly as a function of\u00a0[latex]x[\/latex] by the equation\u00a0[latex]x^{2}+xy-y^{2}+7x-3y-26=0[\/latex]. What is the equation of the tangent line to the graph of this curve at point\u00a0[latex](3, -2)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q995467213\">Show Solution<\/button><\/p>\n<div id=\"q995467213\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\left.\\frac{2x+y+7}{2y-x+3}\\right|_{(3,-2)}=\\frac{2(3)+(-2)+7}{2(-2)-(3)+3}=-\\frac{11}4[\/latex]<\/p>\n<p>Equation of the tangent line:\u00a0[latex]y=-\\frac{11}4x+\\frac{25}4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186162&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=YSpLF3QfzK0&amp;video_target=tpm-plugin-wh05pp33-YSpLF3QfzK0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP4.27_transcript.html\">transcript for \u201cCP 4.27\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":9,"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-618","chapter","type-chapter","status-publish","hentry"],"part":428,"module-header":"fresh_take","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\/618","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\/618\/revisions"}],"predecessor-version":[{"id":1836,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/618\/revisions\/1836"}],"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\/618\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=618"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=618"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=618"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=618"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}