{"id":790,"date":"2025-06-20T17:13:02","date_gmt":"2025-06-20T17:13:02","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=790"},"modified":"2025-08-28T12:58:44","modified_gmt":"2025-08-28T12:58:44","slug":"introduction-to-differential-equations-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/introduction-to-differential-equations-background-youll-need-1\/","title":{"raw":"Introduction to Differential Equations: Background You'll Need 1","rendered":"Introduction to Differential Equations: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Use the chain rule along with other rules to differentiate functions involving powers, products, quotients, and trigonometry<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Combining the Chain Rule With Other Rules<\/h2>\r\n<h3>The Chain and Power Rules Combined<\/h3>\r\n<p id=\"fs-id1169739096228\">When finding derivatives of composite functions, we often need to combine the chain rule with other differentiation rules. For example, to find derivatives of functions of the form [latex]h(x)=(g(x))^n[\/latex], we need to use the chain rule combined with the power rule. To do so, we can think of [latex]h(x)=(g(x))^n[\/latex] as [latex]f(g(x))[\/latex] where [latex]f(x)=x^n[\/latex]. Then [latex]f^{\\prime}(x)=nx^{n-1}[\/latex]. Thus, [latex]f^{\\prime}(g(x))=n(g(x))^{n-1}[\/latex]. This leads us to the derivative of a power function using the chain rule,<\/p>\r\n\r\n<div id=\"fs-id1169739325710\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=n(g(x))^{n-1}g^{\\prime}(x)[\/latex]<\/div>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">power rule for composition of functions<\/h3>\r\n<p id=\"fs-id1169738978364\">For all values of [latex]x[\/latex] for which the derivative is defined, if<\/p>\r\n\r\n<div id=\"fs-id1169739006308\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h(x)=(g(x))^n[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739242349\">Then<\/p>\r\n\r\n<div id=\"fs-id1169739222795\" class=\"equation\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=n(g(x))^{n-1}g^{\\prime}(x)[\/latex]<\/div>\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169736589119\">Find the derivative of [latex]h(x)=\\dfrac{1}{(3x^2+1)^2}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169736658840\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736658840\"]\r\n<p id=\"fs-id1169736658840\">First, rewrite [latex]h(x)=\\frac{1}{(3x^2+1)^2}=(3x^2+1)^{-2}[\/latex].<\/p>\r\n<p id=\"fs-id1169739333152\">Applying the power rule with [latex]g(x)=3x^2+1[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1169736609881\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=-2(3x^2+1)^{-3}(6x)[\/latex].<\/div>\r\n<p id=\"fs-id1169736655793\">Rewriting back to the original form gives us<\/p>\r\n\r\n<div id=\"fs-id1169739008104\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\frac{-12x}{(3x^2+1)^3}[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169739273001\">Find the derivative of [latex]h(x)=\\sin^3 x[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739182335\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739182335\"]\r\n<p id=\"fs-id1169739182335\">First recall that [latex]\\sin^3 x=(\\sin x)^3[\/latex], so we can rewrite [latex]h(x)= \\sin^3 x[\/latex] as [latex]h(x)=(\\sin x)^3[\/latex].<\/p>\r\n<p id=\"fs-id1169739351647\">Applying the power rule with [latex]g(x)= \\sin x[\/latex], we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739039132\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=3(\\sin x)^2 \\cos x=3 \\sin^2 x \\cos x[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169736613825\">Find the equation of a line tangent to the graph of [latex]h(x)=\\dfrac{1}{(3x-5)^2}[\/latex] at [latex]x=2[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169739336066\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739336066\"]\r\n<p id=\"fs-id1169739336066\">Because we are finding an equation of a line, we need a point. The [latex]x[\/latex]-coordinate of the point is 2. To find the [latex]y[\/latex]-coordinate, substitute 2 into [latex]h(x)[\/latex]. Since [latex]h(2)=\\frac{1}{(3(2)-5)^2}=1[\/latex], the point is [latex](2,1)[\/latex].<\/p>\r\n<p id=\"fs-id1169736607681\">For the slope, we need [latex]h^{\\prime}(2)[\/latex]. To find [latex]h^{\\prime}(x)[\/latex], first we rewrite [latex]h(x)=(3x-5)^{-2}[\/latex] and apply the power rule to obtain<\/p>\r\n\r\n<div id=\"fs-id1169736662505\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=-2(3x-5)^{-3}(3)=-6(3x-5)^{-3}[\/latex].<\/div>\r\n<p id=\"fs-id1169739111347\">By substituting, we have [latex]h^{\\prime}(2)=-6(3(2)-5)^{-3}=-6[\/latex]. Therefore, the line has equation [latex]y-1=-6(x-2)[\/latex]. Rewriting, the equation of the line is [latex]y=-6x+13[\/latex].<\/p>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=239&amp;end=341&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.6TheChainRule239to341_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.6 The Chain Rule\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]205911[\/ohm_question]\r\n\r\n<\/section>\r\n<h3>The Chain and Trigonometric Functions Combined<\/h3>\r\n<p id=\"fs-id1169739293763\">Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule.<\/p>\r\n\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739327889\">Find the derivative of [latex]h(x)= \\cos (g(x))[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169736657088\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736657088\"]Think of [latex]h(x)= \\cos (g(x))[\/latex] as [latex]f(g(x))[\/latex] where [latex]f(x)= \\cos x[\/latex]. Since [latex]f^{\\prime}(x)=\u2212\\sin x[\/latex] we have [latex]f^{\\prime}(g(x))=\u2212\\sin (g(x))[\/latex]. Then we do the following calculation.\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}h^{\\prime}(x) &amp; =f^{\\prime}(g(x))g^{\\prime}(x) &amp; &amp; &amp; \\text{Apply the chain rule.} \\\\ &amp; =\u2212\\sin (g(x))g^{\\prime}(x) &amp; &amp; &amp; \\text{Substitute} \\, f^{\\prime}(g(x))=\u2212\\sin (g(x)) \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169739285002\">Thus, the derivative of [latex]h(x)= \\cos (g(x))[\/latex] is given by [latex]h^{\\prime}(x)=\u2212\\sin (g(x))g^{\\prime}(x)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<p id=\"fs-id1169739301534\">In the following example we apply the rule that we have just derived.<\/p>\r\n\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739301547\">Find the derivative of [latex]h(x)= \\cos (5x^2)[\/latex].<\/p>\r\n\r\n[reveal-answer q=\"847788\"]Show Solution[\/reveal-answer] [hidden-answer a=\"847788\"]Let [latex]g(x)=5x^2[\/latex]. Then [latex]g^{\\prime}(x)=10x[\/latex]. Using the result from the previous example,\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}h^{\\prime}(x) &amp; =-\\sin (5x^2) \\cdot 10x \\\\ &amp; =-10x \\sin (5x^2) \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169739333931\">Find the derivative of [latex]h(x)= \\sec (4x^5+2x)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169739300092\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739300092\"]\r\n<p id=\"fs-id1169739300092\">Apply the chain rule to [latex]h(x)= \\sec (g(x))[\/latex] to obtain:<\/p>\r\n\r\n<div id=\"fs-id1169739285070\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)= \\sec (g(x)) \\tan (g(x))g^{\\prime}(x)[\/latex].<\/div>\r\n<p id=\"fs-id1169736615162\">In this problem, [latex]g(x)=4x^5+2x[\/latex], so we have [latex]g^{\\prime}(x)=20x^4+2[\/latex]. Therefore, we obtain,<\/p>\r\n\r\n<div id=\"fs-id1169739299798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}h^{\\prime}(x) &amp; = \\sec (4x^5+2x) \\tan (4x^5+2x)(20x^4+2) \\\\ &amp; =(20x^4+2) \\sec (4x^5+2x) \\tan (4x^5+2x) \\end{array}[\/latex]<\/div>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=411&amp;end=455&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.6TheChainRule411to455_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.6 The Chain Rule\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<p id=\"fs-id1169736610159\">At this stage, we present a collection of derivative formulas derived by applying the chain rule along with the standard derivatives of trigonometric functions. The derivation methods for these formulas are analogous to those demonstrated in the previous examples.<\/p>\r\nFor ease of learning and recall, we have included these formulas in Leibniz's notation, which some students may find more intuitive. Later in this section, we explore the application of the chain rule in Leibniz\u2019s notation in greater detail.\r\n\r\n<section class=\"textbox proTip\">It is important to note that memorizing these formulas as distinct entities is not essential; they are all manifestations of the chain rule applied to well-established derivative formulas.\r\n\r\n<\/section><section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">using the chain rule with trigonometric functions<\/h3>\r\n<p id=\"fs-id1169736655841\">For all values of [latex]x[\/latex] for which the derivative is defined,<\/p>\r\n\r\n<div id=\"fs-id1169736655849\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{llll}\\frac{d}{dx}(\\sin (g(x)))= \\cos (g(x))g^{\\prime}(x) &amp; &amp; &amp; \\frac{d}{dx} \\sin u= \\cos u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\cos (g(x)))=\u2212\\sin (g(x))g^{\\prime}(x) &amp; &amp; &amp; \\frac{d}{dx} \\cos u=\u2212\\sin u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\tan (g(x)))= \\sec^2 (g(x))g^{\\prime}(x) &amp; &amp; &amp; \\frac{d}{dx} \\tan u=\\sec^2 u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\cot (g(x)))=\u2212\\csc^2 (g(x))g^{\\prime}(x) &amp; &amp; &amp; \\frac{d}{dx} \\cot u=\u2212\\csc^2 u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\sec (g(x)))= \\sec (g(x)) \\tan (g(x))g^{\\prime}(x) &amp; &amp; &amp; \\frac{d}{dx} \\sec u= \\sec u \\tan u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\csc (g(x)))=\u2212\\csc (g(x)) \\cot (g(x))g^{\\prime}(x) &amp; &amp; &amp; \\frac{d}{dx} \\csc u=\u2212\\csc u \\cot u\\frac{du}{dx} \\end{array}[\/latex]<\/div>\r\n<\/section>\r\n<h3>The Chain and Product Rules Combined<\/h3>\r\nWhen tackling calculus problems involving products of composite functions, combining the chain and product rules proves indispensable. This approach allows for the systematic differentiation of functions where both rules are necessary to compute the derivative accurately.\r\n\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739298056\">Find the derivative of [latex]h(x)=(2x+1)^5(3x-2)^7[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739345782\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739345782\"]\r\n<p id=\"fs-id1169739345782\">First apply the product rule, then apply the chain rule to each term of the product.<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}h^{\\prime}(x) &amp; =\\frac{d}{dx}((2x+1)^5) \\cdot (3x-2)^7+\\frac{d}{dx}((3x-2)^7) \\cdot (2x+1)^5 &amp; &amp; &amp; \\text{Apply the product rule.} \\\\ &amp; =5(2x+1)^4 \\cdot 2 \\cdot (3x-2)^7+7(3x-2)^6 \\cdot 3 \\cdot (2x+1)^5 &amp; &amp; &amp; \\text{Apply the chain rule.} \\\\ &amp; =10(2x+1)^4(3x-2)^7+21(3x-2)^6(2x+1)^5 &amp; &amp; &amp; \\text{Simplify.} \\\\ &amp; =(2x+1)^4(3x-2)^6(10(3x-2)+21(2x+1)) &amp; &amp; &amp; \\text{Factor out} \\, (2x+1)^4(3x-2)^6. \\\\ &amp; =(2x+1)^4(3x-2)^6(72x+1) &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=457&amp;end=630&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.6TheChainRule457to630_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.6 The Chain Rule\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]206007[\/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\">Use the chain rule along with other rules to differentiate functions involving powers, products, quotients, and trigonometry<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Combining the Chain Rule With Other Rules<\/h2>\n<h3>The Chain and Power Rules Combined<\/h3>\n<p id=\"fs-id1169739096228\">When finding derivatives of composite functions, we often need to combine the chain rule with other differentiation rules. For example, to find derivatives of functions of the form [latex]h(x)=(g(x))^n[\/latex], we need to use the chain rule combined with the power rule. To do so, we can think of [latex]h(x)=(g(x))^n[\/latex] as [latex]f(g(x))[\/latex] where [latex]f(x)=x^n[\/latex]. Then [latex]f^{\\prime}(x)=nx^{n-1}[\/latex]. Thus, [latex]f^{\\prime}(g(x))=n(g(x))^{n-1}[\/latex]. This leads us to the derivative of a power function using the chain rule,<\/p>\n<div id=\"fs-id1169739325710\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=n(g(x))^{n-1}g^{\\prime}(x)[\/latex]<\/div>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">power rule for composition of functions<\/h3>\n<p id=\"fs-id1169738978364\">For all values of [latex]x[\/latex] for which the derivative is defined, if<\/p>\n<div id=\"fs-id1169739006308\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h(x)=(g(x))^n[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739242349\">Then<\/p>\n<div id=\"fs-id1169739222795\" class=\"equation\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=n(g(x))^{n-1}g^{\\prime}(x)[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169736589119\">Find the derivative of [latex]h(x)=\\dfrac{1}{(3x^2+1)^2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169736658840\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169736658840\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736658840\">First, rewrite [latex]h(x)=\\frac{1}{(3x^2+1)^2}=(3x^2+1)^{-2}[\/latex].<\/p>\n<p id=\"fs-id1169739333152\">Applying the power rule with [latex]g(x)=3x^2+1[\/latex], we have<\/p>\n<div id=\"fs-id1169736609881\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=-2(3x^2+1)^{-3}(6x)[\/latex].<\/div>\n<p id=\"fs-id1169736655793\">Rewriting back to the original form gives us<\/p>\n<div id=\"fs-id1169739008104\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\frac{-12x}{(3x^2+1)^3}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739273001\">Find the derivative of [latex]h(x)=\\sin^3 x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739182335\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739182335\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739182335\">First recall that [latex]\\sin^3 x=(\\sin x)^3[\/latex], so we can rewrite [latex]h(x)= \\sin^3 x[\/latex] as [latex]h(x)=(\\sin x)^3[\/latex].<\/p>\n<p id=\"fs-id1169739351647\">Applying the power rule with [latex]g(x)= \\sin x[\/latex], we obtain<\/p>\n<div id=\"fs-id1169739039132\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=3(\\sin x)^2 \\cos x=3 \\sin^2 x \\cos x[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169736613825\">Find the equation of a line tangent to the graph of [latex]h(x)=\\dfrac{1}{(3x-5)^2}[\/latex] at [latex]x=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739336066\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739336066\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739336066\">Because we are finding an equation of a line, we need a point. The [latex]x[\/latex]-coordinate of the point is 2. To find the [latex]y[\/latex]-coordinate, substitute 2 into [latex]h(x)[\/latex]. Since [latex]h(2)=\\frac{1}{(3(2)-5)^2}=1[\/latex], the point is [latex](2,1)[\/latex].<\/p>\n<p id=\"fs-id1169736607681\">For the slope, we need [latex]h^{\\prime}(2)[\/latex]. To find [latex]h^{\\prime}(x)[\/latex], first we rewrite [latex]h(x)=(3x-5)^{-2}[\/latex] and apply the power rule to obtain<\/p>\n<div id=\"fs-id1169736662505\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=-2(3x-5)^{-3}(3)=-6(3x-5)^{-3}[\/latex].<\/div>\n<p id=\"fs-id1169739111347\">By substituting, we have [latex]h^{\\prime}(2)=-6(3(2)-5)^{-3}=-6[\/latex]. Therefore, the line has equation [latex]y-1=-6(x-2)[\/latex]. Rewriting, the equation of the line is [latex]y=-6x+13[\/latex].<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=239&amp;end=341&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/button><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.6TheChainRule239to341_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.6 The Chain Rule&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm205911\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=205911&theme=lumen&iframe_resize_id=ohm205911&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<h3>The Chain and Trigonometric Functions Combined<\/h3>\n<p id=\"fs-id1169739293763\">Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739327889\">Find the derivative of [latex]h(x)= \\cos (g(x))[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169736657088\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169736657088\" class=\"hidden-answer\" style=\"display: none\">Think of [latex]h(x)= \\cos (g(x))[\/latex] as [latex]f(g(x))[\/latex] where [latex]f(x)= \\cos x[\/latex]. Since [latex]f^{\\prime}(x)=\u2212\\sin x[\/latex] we have [latex]f^{\\prime}(g(x))=\u2212\\sin (g(x))[\/latex]. Then we do the following calculation.<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}h^{\\prime}(x) & =f^{\\prime}(g(x))g^{\\prime}(x) & & & \\text{Apply the chain rule.} \\\\ & =\u2212\\sin (g(x))g^{\\prime}(x) & & & \\text{Substitute} \\, f^{\\prime}(g(x))=\u2212\\sin (g(x)) \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169739285002\">Thus, the derivative of [latex]h(x)= \\cos (g(x))[\/latex] is given by [latex]h^{\\prime}(x)=\u2212\\sin (g(x))g^{\\prime}(x)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1169739301534\">In the following example we apply the rule that we have just derived.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739301547\">Find the derivative of [latex]h(x)= \\cos (5x^2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q847788\">Show Solution<\/button> <\/p>\n<div id=\"q847788\" class=\"hidden-answer\" style=\"display: none\">Let [latex]g(x)=5x^2[\/latex]. Then [latex]g^{\\prime}(x)=10x[\/latex]. Using the result from the previous example,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}h^{\\prime}(x) & =-\\sin (5x^2) \\cdot 10x \\\\ & =-10x \\sin (5x^2) \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739333931\">Find the derivative of [latex]h(x)= \\sec (4x^5+2x)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739300092\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739300092\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739300092\">Apply the chain rule to [latex]h(x)= \\sec (g(x))[\/latex] to obtain:<\/p>\n<div id=\"fs-id1169739285070\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)= \\sec (g(x)) \\tan (g(x))g^{\\prime}(x)[\/latex].<\/div>\n<p id=\"fs-id1169736615162\">In this problem, [latex]g(x)=4x^5+2x[\/latex], so we have [latex]g^{\\prime}(x)=20x^4+2[\/latex]. Therefore, we obtain,<\/p>\n<div id=\"fs-id1169739299798\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}h^{\\prime}(x) & = \\sec (4x^5+2x) \\tan (4x^5+2x)(20x^4+2) \\\\ & =(20x^4+2) \\sec (4x^5+2x) \\tan (4x^5+2x) \\end{array}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=411&amp;end=455&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.6TheChainRule411to455_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.6 The Chain Rule&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1169736610159\">At this stage, we present a collection of derivative formulas derived by applying the chain rule along with the standard derivatives of trigonometric functions. The derivation methods for these formulas are analogous to those demonstrated in the previous examples.<\/p>\n<p>For ease of learning and recall, we have included these formulas in Leibniz&#8217;s notation, which some students may find more intuitive. Later in this section, we explore the application of the chain rule in Leibniz\u2019s notation in greater detail.<\/p>\n<section class=\"textbox proTip\">It is important to note that memorizing these formulas as distinct entities is not essential; they are all manifestations of the chain rule applied to well-established derivative formulas.<\/p>\n<\/section>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">using the chain rule with trigonometric functions<\/h3>\n<p id=\"fs-id1169736655841\">For all values of [latex]x[\/latex] for which the derivative is defined,<\/p>\n<div id=\"fs-id1169736655849\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{llll}\\frac{d}{dx}(\\sin (g(x)))= \\cos (g(x))g^{\\prime}(x) & & & \\frac{d}{dx} \\sin u= \\cos u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\cos (g(x)))=\u2212\\sin (g(x))g^{\\prime}(x) & & & \\frac{d}{dx} \\cos u=\u2212\\sin u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\tan (g(x)))= \\sec^2 (g(x))g^{\\prime}(x) & & & \\frac{d}{dx} \\tan u=\\sec^2 u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\cot (g(x)))=\u2212\\csc^2 (g(x))g^{\\prime}(x) & & & \\frac{d}{dx} \\cot u=\u2212\\csc^2 u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\sec (g(x)))= \\sec (g(x)) \\tan (g(x))g^{\\prime}(x) & & & \\frac{d}{dx} \\sec u= \\sec u \\tan u\\frac{du}{dx} \\\\ \\frac{d}{dx}(\\csc (g(x)))=\u2212\\csc (g(x)) \\cot (g(x))g^{\\prime}(x) & & & \\frac{d}{dx} \\csc u=\u2212\\csc u \\cot u\\frac{du}{dx} \\end{array}[\/latex]<\/div>\n<\/section>\n<h3>The Chain and Product Rules Combined<\/h3>\n<p>When tackling calculus problems involving products of composite functions, combining the chain and product rules proves indispensable. This approach allows for the systematic differentiation of functions where both rules are necessary to compute the derivative accurately.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739298056\">Find the derivative of [latex]h(x)=(2x+1)^5(3x-2)^7[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739345782\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739345782\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739345782\">First apply the product rule, then apply the chain rule to each term of the product.<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}h^{\\prime}(x) & =\\frac{d}{dx}((2x+1)^5) \\cdot (3x-2)^7+\\frac{d}{dx}((3x-2)^7) \\cdot (2x+1)^5 & & & \\text{Apply the product rule.} \\\\ & =5(2x+1)^4 \\cdot 2 \\cdot (3x-2)^7+7(3x-2)^6 \\cdot 3 \\cdot (2x+1)^5 & & & \\text{Apply the chain rule.} \\\\ & =10(2x+1)^4(3x-2)^7+21(3x-2)^6(2x+1)^5 & & & \\text{Simplify.} \\\\ & =(2x+1)^4(3x-2)^6(10(3x-2)+21(2x+1)) & & & \\text{Factor out} \\, (2x+1)^4(3x-2)^6. \\\\ & =(2x+1)^4(3x-2)^6(72x+1) & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=457&amp;end=630&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.6TheChainRule457to630_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.6 The Chain Rule&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm206007\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206007&theme=lumen&iframe_resize_id=ohm206007&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":669,"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\/790"}],"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":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/790\/revisions"}],"predecessor-version":[{"id":2037,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/790\/revisions\/2037"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/669"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/790\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=790"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=790"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=790"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=790"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}