{"id":976,"date":"2025-06-20T17:26:23","date_gmt":"2025-06-20T17:26:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=976"},"modified":"2025-08-28T13:35:17","modified_gmt":"2025-08-28T13:35:17","slug":"parametric-curves-and-their-applications-background-youll-need-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/parametric-curves-and-their-applications-background-youll-need-3\/","title":{"raw":"Parametric Curves and Their Applications: Background You'll Need 3","rendered":"Parametric Curves and Their Applications: Background You&#8217;ll Need 3"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Explain and use the chain rule<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Chain Rule<\/h2>\r\n<p id=\"fs-id1169738824611\">The <strong>chain rule<\/strong> simplifies the process of differentiating composite functions by breaking down a function into simpler parts. Rather than differentiating the entire composite directly, the chain rule allows us to differentiate each part independently. Specifically, for a composite function [latex]h(x)=f(g(x))[\/latex], the derivative is given by the derivative of the outer function [latex]f[\/latex] evaluated at the inner function [latex]g(x)[\/latex], multiplied by the derivative of the inner function:<\/p>\r\n<p style=\"text-align: center;\">[latex]h^{\\prime} (x)=f^{\\prime} (g(x)) \\cdot g^{\\prime} (x)[\/latex]<\/p>\r\n\r\n<section class=\"textbox example\">To contextualize this, consider [latex]h(x)= \\sin (x^3)[\/latex].\r\n\r\nFirst, identify the inner function [latex]g(x)=x^3[\/latex] and the outer function[latex] f(u)= \\sin u[\/latex], where [latex]u=g(x)[\/latex]. Applying the chain rule, which states the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function, we proceed as follows:\r\n<ol>\r\n \t<li>Differentiate the inner function [latex]g(x)[\/latex]:<center>[latex]g^{\\prime} (x) = 3x^2[\/latex]<\/center><\/li>\r\n \t<li>Differentiate the outer function [latex]f(u) [\/latex] with respect to [latex]u[\/latex]:<center>[latex]f^{\\prime}(u) = \\cos u[\/latex]<\/center><\/li>\r\n \t<li>Since [latex]u=g(x)=x^3[\/latex], substituting that back in, we get:<center>[latex]f^{\\prime} (u) = \\cos{(x^3)}[\/latex]<\/center><\/li>\r\n \t<li>Apply the chain rule:<center>[latex]h^{\\prime} (x)=f^{\\prime} (g(x)) \\cdot g^{\\prime} (x)= \\cos{(x^3)} \\cdot 3x^2[\/latex]<\/center><\/li>\r\n<\/ol>\r\nTherefore, the derivative of [latex]h(x)= \\sin (x^3)[\/latex] is:<center>[latex] h^{\\prime} (x)= 3x^2\\cos{(x^3)}[\/latex]<\/center><\/section>\r\n<p id=\"fs-id1169738955406\">Now that we've illustrated how to apply the chain rule with a specific example, let's explore the general formula of the chain rule and see how it applies to various types of composite functions. An informal proof of this concept will follow at the end of this section.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">the chain rule<\/h3>\r\n<p id=\"fs-id1169736612521\">Let [latex]f[\/latex] and [latex]g[\/latex] be functions. For all [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g[\/latex] is differentiable at [latex]x[\/latex] and [latex]f[\/latex] is differentiable at [latex]g(x)[\/latex], the derivative of the composite function [latex]h(x)=(f\\circ g)(x)=f(g(x))[\/latex]\u00a0is given by:<\/p>\r\n\r\n<div id=\"fs-id1169738948938\" class=\"equation\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=f^{\\prime}(g(x)) \\cdot g^{\\prime}(x)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739199751\">Alternatively, if [latex]y[\/latex] is a function of [latex]u[\/latex], and [latex]u[\/latex] is a function of [latex]x[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1169739187558\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{dy}{du} \\cdot \\frac{du}{dx}[\/latex]<\/div>\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]288385[\/ohm_question]\r\n\r\n<\/section><section class=\"textbox interact\">For more information, check out this <a href=\"http:\/\/webspace.ship.edu\/msrenault\/GeoGebraCalculus\/derivative_intuitive_chain_rule.html\" target=\"_blank\" rel=\"noopener\">interactive on The Intuitive Notion of the Chain Rule.<\/a>\r\n\r\n<\/section><section class=\"textbox questionHelp\"><strong>How to: Apply the Chain Rule<\/strong>\r\n<ol id=\"fs-id1169739182671\">\r\n \t<li><strong>Identify the Functions:<\/strong> Begin by identifying the inner function [latex]g(x)[\/latex] and the outer function [latex]f(u)[\/latex], where [latex]u=g(x)[\/latex].<\/li>\r\n \t<li><strong>Derivative of the Outer Function:<\/strong> Compute the derivative of the outer function, [latex]f^{\\prime}(u)[\/latex], and evaluate it at [latex]g(x)[\/latex] to obtain [latex]f^{\\prime}(g(x))[\/latex].<\/li>\r\n \t<li><strong>Derivative of the Inner Function:<\/strong> Calculate the derivative of the inner function [latex]g^{\\prime}(x)[\/latex].<\/li>\r\n \t<li><strong>Apply the Chain Rule:<\/strong> Combine these results to find [latex]h^{\\prime}(x)[\/latex] as follows:\u00a0<center>[latex]h^{\\prime}(x)=f^{\\prime}(g(x)) \\cdot g^{\\prime}(x)[\/latex]<\/center><\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox proTip\">When applying the chain rule to compositions involving multiple functions, remember that each function contributes to the derivative. Start from the innermost function and work outward, applying the chain rule iteratively. The derivative of a composite involving three functions, for example, involves taking derivatives step by step, moving from the innermost to the outermost function. Importantly, derivatives are not evaluated at derivatives; they are evaluated at functions.\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]206069[\/ohm_question]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Explain and use the chain rule<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>The Chain Rule<\/h2>\n<p id=\"fs-id1169738824611\">The <strong>chain rule<\/strong> simplifies the process of differentiating composite functions by breaking down a function into simpler parts. Rather than differentiating the entire composite directly, the chain rule allows us to differentiate each part independently. Specifically, for a composite function [latex]h(x)=f(g(x))[\/latex], the derivative is given by the derivative of the outer function [latex]f[\/latex] evaluated at the inner function [latex]g(x)[\/latex], multiplied by the derivative of the inner function:<\/p>\n<p style=\"text-align: center;\">[latex]h^{\\prime} (x)=f^{\\prime} (g(x)) \\cdot g^{\\prime} (x)[\/latex]<\/p>\n<section class=\"textbox example\">To contextualize this, consider [latex]h(x)= \\sin (x^3)[\/latex].<\/p>\n<p>First, identify the inner function [latex]g(x)=x^3[\/latex] and the outer function[latex]f(u)= \\sin u[\/latex], where [latex]u=g(x)[\/latex]. Applying the chain rule, which states the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function, we proceed as follows:<\/p>\n<ol>\n<li>Differentiate the inner function [latex]g(x)[\/latex]:\n<div style=\"text-align: center;\">[latex]g^{\\prime} (x) = 3x^2[\/latex]<\/div>\n<\/li>\n<li>Differentiate the outer function [latex]f(u)[\/latex] with respect to [latex]u[\/latex]:\n<div style=\"text-align: center;\">[latex]f^{\\prime}(u) = \\cos u[\/latex]<\/div>\n<\/li>\n<li>Since [latex]u=g(x)=x^3[\/latex], substituting that back in, we get:\n<div style=\"text-align: center;\">[latex]f^{\\prime} (u) = \\cos{(x^3)}[\/latex]<\/div>\n<\/li>\n<li>Apply the chain rule:\n<div style=\"text-align: center;\">[latex]h^{\\prime} (x)=f^{\\prime} (g(x)) \\cdot g^{\\prime} (x)= \\cos{(x^3)} \\cdot 3x^2[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p>Therefore, the derivative of [latex]h(x)= \\sin (x^3)[\/latex] is:<\/p>\n<div style=\"text-align: center;\">[latex]h^{\\prime} (x)= 3x^2\\cos{(x^3)}[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1169738955406\">Now that we&#8217;ve illustrated how to apply the chain rule with a specific example, let&#8217;s explore the general formula of the chain rule and see how it applies to various types of composite functions. An informal proof of this concept will follow at the end of this section.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">the chain rule<\/h3>\n<p id=\"fs-id1169736612521\">Let [latex]f[\/latex] and [latex]g[\/latex] be functions. For all [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g[\/latex] is differentiable at [latex]x[\/latex] and [latex]f[\/latex] is differentiable at [latex]g(x)[\/latex], the derivative of the composite function [latex]h(x)=(f\\circ g)(x)=f(g(x))[\/latex]\u00a0is given by:<\/p>\n<div id=\"fs-id1169738948938\" class=\"equation\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=f^{\\prime}(g(x)) \\cdot g^{\\prime}(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739199751\">Alternatively, if [latex]y[\/latex] is a function of [latex]u[\/latex], and [latex]u[\/latex] is a function of [latex]x[\/latex], then<\/p>\n<div id=\"fs-id1169739187558\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{dy}{du} \\cdot \\frac{du}{dx}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm288385\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288385&theme=lumen&iframe_resize_id=ohm288385&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<section class=\"textbox interact\">For more information, check out this <a href=\"http:\/\/webspace.ship.edu\/msrenault\/GeoGebraCalculus\/derivative_intuitive_chain_rule.html\" target=\"_blank\" rel=\"noopener\">interactive on The Intuitive Notion of the Chain Rule.<\/a><\/p>\n<\/section>\n<section class=\"textbox questionHelp\"><strong>How to: Apply the Chain Rule<\/strong><\/p>\n<ol id=\"fs-id1169739182671\">\n<li><strong>Identify the Functions:<\/strong> Begin by identifying the inner function [latex]g(x)[\/latex] and the outer function [latex]f(u)[\/latex], where [latex]u=g(x)[\/latex].<\/li>\n<li><strong>Derivative of the Outer Function:<\/strong> Compute the derivative of the outer function, [latex]f^{\\prime}(u)[\/latex], and evaluate it at [latex]g(x)[\/latex] to obtain [latex]f^{\\prime}(g(x))[\/latex].<\/li>\n<li><strong>Derivative of the Inner Function:<\/strong> Calculate the derivative of the inner function [latex]g^{\\prime}(x)[\/latex].<\/li>\n<li><strong>Apply the Chain Rule:<\/strong> Combine these results to find [latex]h^{\\prime}(x)[\/latex] as follows:\u00a0\n<div style=\"text-align: center;\">[latex]h^{\\prime}(x)=f^{\\prime}(g(x)) \\cdot g^{\\prime}(x)[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox proTip\">When applying the chain rule to compositions involving multiple functions, remember that each function contributes to the derivative. Start from the innermost function and work outward, applying the chain rule iteratively. The derivative of a composite involving three functions, for example, involves taking derivatives step by step, moving from the innermost to the outermost function. Importantly, derivatives are not evaluated at derivatives; they are evaluated at functions.<\/p>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm206069\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206069&theme=lumen&iframe_resize_id=ohm206069&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":675,"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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/976"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":2,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/976\/revisions"}],"predecessor-version":[{"id":2059,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/976\/revisions\/2059"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/675"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/976\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=976"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=976"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=976"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=976"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}