{"id":349,"date":"2021-02-04T01:18:14","date_gmt":"2021-02-04T01:18:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=349"},"modified":"2022-03-16T05:34:41","modified_gmt":"2022-03-16T05:34:41","slug":"combining-the-chain-rule-with-other-rules","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/combining-the-chain-rule-with-other-rules\/","title":{"raw":"Combining the Chain Rule With Other Rules","rendered":"Combining the Chain Rule With Other Rules"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply the chain rule together with the power rule<\/li>\r\n \t<li>Apply the chain rule and the product\/quotient rules correctly in combination when both are necessary<\/li>\r\n \t<li>Recognize the chain rule for a composition of three or more functions<\/li>\r\n \t<li>Describe the proof of the chain rule<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Chain and Power Rules Combined<\/h2>\r\n<p id=\"fs-id1169739096228\">We can now apply the chain rule to composite functions, but note that we often need to use it with other 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&nbsp;\r\n<div><\/div>\r\n<div id=\"fs-id1169739187734\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Power Rule for Composition of Functions<\/h3>\r\n\r\n<hr \/>\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&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739274312\" class=\"textbook exercises\">\r\n<h3>Example: Using the Chain and Power Rules<\/h3>\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&nbsp;\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<\/div>\r\n<div id=\"fs-id1169736662938\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169736595961\">Find the derivative of [latex]h(x)=(2x^3+2x-1)^4[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739325717\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739325717\"]\r\n<p id=\"fs-id1169739325717\">[latex]h^{\\prime}(x)=4(2x^3+2x-1)^3(6x^{2}+2)=8(3x^{2}+1)(2x^3+2x-1)^3[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1169739274677\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169739179049\">Use the previous example with [latex]g(x)=2x^3+2x-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=136&amp;end=185&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]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\/3.6TheChainRule136to185_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.6 The Chain Rule\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169739302258\" class=\"textbook exercises\">\r\n<h3>Example: Using the Chain and Power Rules with a Trigonometric Function<\/h3>\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\">[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<\/div>\r\n<div id=\"fs-id1169739190027\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Equation of a Tangent Line<\/h3>\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&nbsp;\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\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding the Equation of a Tangent Line.\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\"><\/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>.[\/hidden-answer]\r\n<div id=\"fs-id1169739301868\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739303885\">Find the equation of the line tangent to the graph of [latex]f(x)=(x^2-2)^3[\/latex] at [latex]x=-2[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169738869705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738869705\"]\r\n<p id=\"fs-id1169738869705\">[latex]y=-48x-88[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169739252012\">Use the preceding example as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]205911[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>The Chain and Trigonometric Functions Combined<\/h2>\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<div id=\"fs-id1169736656617\" class=\"textbook exercises\">\r\n<h3>Example: Using the Chain Rule on a General Cosine Function<\/h3>\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<div><\/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<\/div>\r\n<p id=\"fs-id1169739301534\">In the following example we apply the rule that we have just derived.<\/p>\r\n\r\n<div id=\"fs-id1169739301537\" class=\"textbook exercises\">\r\n<h3>Example: Using the Chain Rule on a Cosine Function<\/h3>\r\n<p id=\"fs-id1169739301547\">Find the derivative of [latex]h(x)= \\cos (5x^2)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169739111373\" class=\"equation unnumbered\">[reveal-answer q=\"847788\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847788\"]Let [latex]g(x)=5x^2[\/latex]. Then [latex]g^{\\prime}(x)=10x[\/latex].\r\nUsing the result from the previous example, [latex]\\begin{array}{ll}h^{\\prime}(x) &amp; =-\\sin (5x^2) \\cdot 10x \\\\ &amp; =-10x \\sin (5x^2) \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739333921\" class=\"textbook exercises\">\r\n<h3>Example: Using the Chain Rule on Another Trigonometric Function<\/h3>\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\">[latex]h^{\\prime}(x)= \\sec (g(x)) \\tan (g(x))g^{\\prime}(x)[\/latex].<\/div>\r\n&nbsp;\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\">[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\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using the Chain Rule on Another Trigonometric Function.\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\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]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>.[\/hidden-answer]\r\n<div id=\"fs-id1169739188144\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739188151\">Find the derivative of [latex]h(x)= \\sin (7x+2)[\/latex].<\/p>\r\n[reveal-answer q=\"166577\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"166577\"]\r\n<p id=\"fs-id1169736607578\">Apply the chain rule to [latex]h(x)= \\sin g(x)[\/latex] first and then use [latex]g(x)=7x+2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"232193\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"232193\"]\r\n\r\n[latex]h^{\\prime}(x)=7 \\cos (7x+2)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1169736610159\">At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in the last three examples. For convenience, formulas are also given in Leibniz\u2019s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz\u2019s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.<\/p>\r\n\r\n<div id=\"fs-id1169736610174\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Using the Chain Rule with Trigonometric Functions<\/h3>\r\n\r\n<hr \/>\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&nbsp;\r\n\r\n<\/div>\r\n<h2>The Chain and Product Rules Combined<\/h2>\r\n<div id=\"fs-id1169739298047\" class=\"textbook exercises\">\r\n<h3>Example: Combining the Chain Rule with the Product Rule<\/h3>\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\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Combining the Chain Rule with the Product Rule.\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\"><\/iframe><\/center>[reveal-answer q=\"266836\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266836\"]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>.[\/hidden-answer]\r\n<div id=\"fs-id1169736603462\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169736603469\">Find the derivative of [latex]h(x)=\\dfrac{x}{(2x+3)^3}[\/latex]<\/p>\r\n[reveal-answer q=\"3778210\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3778210\"]\r\n<p id=\"fs-id1169736603573\">Start out by applying the quotient rule. Remember to use the chain rule to differentiate the denominator.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169736603515\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736603515\"]\r\n<p id=\"fs-id1169736603515\">[latex]h^{\\prime}(x)=\\frac{3-4x}{(2x+3)^4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]206007[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Applying the Chain Rule Multiple Times<\/h2>\r\n<p id=\"fs-id1169739304867\">We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple times.<\/p>\r\n<p id=\"fs-id1169739304874\">In general terms, first we let<\/p>\r\n\r\n<div id=\"fs-id1169739304877\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k(x)=h(f(g(x)))[\/latex]<\/div>\r\nThen, applying the chain rule once we obtain\r\n<div id=\"fs-id1169739304927\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=\\frac{d}{dx}(h(f(g(x))))=h^{\\prime}(f(g(x))) \\cdot \\frac{d}{dx}(f(g(x)))[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736597613\">Applying the chain rule again, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169736597616\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=h^{\\prime}(f(g(x)))f^{\\prime}(g(x))g^{\\prime}(x)[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div id=\"fs-id1169736597703\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Chain Rule for a Composition of Three Functions<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169736597709\">For all values of [latex]x[\/latex] for which the function is differentiable, if<\/p>\r\n\r\n<div id=\"fs-id1169736597716\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k(x)=h(f(g(x)))[\/latex],<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1169736597763\">then<\/p>\r\n\r\n<div id=\"fs-id1169736597766\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=h^{\\prime}(f(g(x)))f^{\\prime}(g(x))g^{\\prime}(x)[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1169736658568\">In other words, we are applying the chain rule twice.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1169736658572\">Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, <em>remember, we always work from the outside in, taking one derivative at a time.<\/em><\/p>\r\n\r\n<div id=\"fs-id1169736658580\" class=\"textbook exercises\">\r\n<h3>Example: Differentiating a Composite of Three Functions<\/h3>\r\n<p id=\"fs-id1169736658589\">Find the derivative of [latex]k(x)=\\cos^4 (7x^2+1)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169736658640\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736658640\"]\r\n<p id=\"fs-id1169736658640\">First, rewrite [latex]k(x)[\/latex] as<\/p>\r\n\r\n<div id=\"fs-id1169736658656\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k(x)=(\\cos(7x^2+1))^4[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739275169\">Then apply the chain rule several times.<\/p>\r\n\r\n<div id=\"fs-id1169739275173\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}k^{\\prime}(x) &amp; =4(\\cos(7x^2+1))^3(\\frac{d}{dx}\\cos(7x^2+1)) &amp; &amp; &amp; \\text{Apply the chain rule.} \\\\ &amp; =4(\\cos(7x^2+1))^3(\u2212\\sin(7x^2+1))(\\frac{d}{dx}(7x^2+1)) &amp; &amp; &amp; \\text{Apply the chain rule.} \\\\ &amp; =4(\\cos(7x^2+1))^3(\u2212\\sin(7x^2+1))(14x) &amp; &amp; &amp; \\text{Apply the chain rule.} \\\\ &amp; =-56x \\sin(7x^2+1) \\cos^3 (7x^2+1) &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div>Don't forget that\u00a0[latex]{\\cos }^{n}t[\/latex] is a commonly used shorthand notation for [latex]{\\left(\\cos \\left(t\\right)\\right)}^{n}[\/latex]. When we write it without the shorthand notation, we can clearly see why the chain rule is necessary in these situations.<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1169739350744\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739350751\">Find the derivative of [latex]h(x)=\\sin^6 (x^3)[\/latex]<\/p>\r\n[reveal-answer q=\"9044612\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"9044612\"]\r\n<p id=\"fs-id1169739350858\">Rewrite [latex]h(x)=\\sin^6 (x^3)=(\\sin(x^3))^6[\/latex] and use the previous example as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739350792\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739350792\"]\r\n<p id=\"fs-id1169739350792\">[latex]h^{\\prime}(x)=18x^2 \\sin^5 (x^3) \\cos (x^3)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]206008[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169736593542\" class=\"textbook exercises\">\r\n<h3>Example: Using the Chain Rule in a Velocity Problem<\/h3>\r\n<p id=\"fs-id1169736593552\">A particle moves along a coordinate axis. Its position at time [latex]t[\/latex] is given by [latex]s(t)= \\sin (2t)+ \\cos (3t)[\/latex]. What is the velocity of the particle at time [latex]t=\\frac{\\pi}{6}[\/latex]?<\/p>\r\n[reveal-answer q=\"fs-id1169736593621\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736593621\"]\r\n<p id=\"fs-id1169736593621\">To find [latex]v(t)[\/latex], the velocity of the particle at time [latex]t[\/latex], we must differentiate [latex]s(t)[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1169736593661\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=s^{\\prime}(t)=2 \\cos(2t)-3 \\sin(3t)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739266652\">Substituting [latex]t=\\frac{\\pi}{6}[\/latex] into [latex]v(t)[\/latex], we obtain [latex]v(\\frac{\\pi}{6})=-2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using the Chain Rule in a Velocity Problem.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=725&amp;end=835&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266837\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266837\"]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\/3.6TheChainRule725to835_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.6 The Chain Rule\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169739266711\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739266719\">A particle moves along a coordinate axis. Its position at time [latex]t[\/latex] is given by [latex]s(t)= \\sin (4t)[\/latex]. Find its acceleration at time [latex]t[\/latex].<\/p>\r\n[reveal-answer q=\"3388209\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3388209\"]\r\n<p id=\"fs-id1169736655167\">Acceleration is the second derivative of position.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739266765\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739266765\"]\r\n<p id=\"fs-id1169739266765\">[latex]a(t)=-16 \\sin (4t)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Proof of the Chain Rule<\/h3>\r\n<p id=\"fs-id1169736655179\">At this point, we present a very informal proof of the chain rule. For simplicity\u2019s sake we ignore certain issues: For example, we assume that [latex]g(x)\\ne g(a)[\/latex] for [latex]x\\ne a[\/latex] in some open interval containing [latex]a[\/latex]. We begin by applying the limit definition of the derivative to the function [latex]h(x)[\/latex] to obtain [latex]h^{\\prime}(a)[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1169736655258\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{x-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739305452\">Rewriting, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739305455\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)} \\cdot \\dfrac{g(x)-g(a)}{x-a}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739305587\">Although it is clear that<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{g(x)-g(a)}{x-a}=g^{\\prime}(a)[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736662280\">it is not obvious that<\/p>\r\n\r\n<div id=\"fs-id1169736662283\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)}=f^{\\prime}(g(a))[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736662391\">To see that this is true, first recall that since [latex]g[\/latex] is differentiable at [latex]a, \\, g[\/latex] is also continuous at [latex]a[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1169736662414\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}g(x)=g(a)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739303393\">Next, make the substitution [latex]y=g(x)[\/latex] and [latex]b=g(a)[\/latex] and use change of variables in the limit to obtain<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)}=\\underset{y\\to b}{\\lim}\\dfrac{f(y)-f(b)}{y-b}=f^{\\prime}(b)=f^{\\prime}(g(a))[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739335832\">Finally,<\/p>\r\n\r\n<div id=\"fs-id1169739335835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)} \\cdot \\dfrac{g(x)-g(a)}{x-a}=f^{\\prime}(g(a))g^{\\prime}(a)[\/latex]<\/div>\r\n[latex]_\\blacksquare[\/latex]\r\n<div><\/div>\r\n<div id=\"fs-id1169736613849\" class=\"textbook exercises\">\r\n<h3>Example: Using the Chain Rule with Functional Values<\/h3>\r\n<p id=\"fs-id1169736613859\">Let [latex]h(x)=f(g(x))[\/latex]. If [latex]g(1)=4, \\, g^{\\prime}(1)=3[\/latex], and [latex]f^{\\prime}(4)=7[\/latex], find [latex]h^{\\prime}(1)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169736613978\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736613978\"]\r\n<p id=\"fs-id1169736613978\">Use the chain rule, then substitute.<\/p>\r\n\r\n<div id=\"fs-id1169736613982\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}h^{\\prime}(1) &amp; =f^{\\prime}(g(1))g^{\\prime}(1) &amp; &amp; &amp; \\text{Apply the chain rule.} \\\\ &amp; =f^{\\prime}(4) \\cdot 3 &amp; &amp; &amp; \\text{Substitute} \\, g(1)=4 \\, \\text{and} \\, g^{\\prime}(1)=3. \\\\ &amp; =7 \\cdot 3 &amp; &amp; &amp; \\text{Substitute} \\, f^{\\prime}(4)=7. \\\\ &amp; =21 &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using the Chain Rule with Functional Values.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=842&amp;end=920&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266838\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266838\"]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\/3.6TheChainRule842to920_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.6 The Chain Rule\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169739341441\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739341448\">Let [latex]h(x)=f(g(x))[\/latex]. If [latex]g(2)=-3, \\, g^{\\prime}(2)=4[\/latex], and [latex]f^{\\prime}(-3)=7[\/latex], find [latex]h^{\\prime}(2)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169739353312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739353312\"]\r\n<p id=\"fs-id1169739353312\">28<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169739353326\">Follow the example of using the chain rule in a velocity problem.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply the chain rule together with the power rule<\/li>\n<li>Apply the chain rule and the product\/quotient rules correctly in combination when both are necessary<\/li>\n<li>Recognize the chain rule for a composition of three or more functions<\/li>\n<li>Describe the proof of the chain rule<\/li>\n<\/ul>\n<\/div>\n<h2>The Chain and Power Rules Combined<\/h2>\n<p id=\"fs-id1169739096228\">We can now apply the chain rule to composite functions, but note that we often need to use it with other 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<p>&nbsp;<\/p>\n<div><\/div>\n<div id=\"fs-id1169739187734\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Power Rule for Composition of Functions<\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1169739274312\" class=\"textbook exercises\">\n<h3>Example: Using the Chain and Power Rules<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736658840\">Show Solution<\/span><\/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>&nbsp;<\/p>\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<\/div>\n<div id=\"fs-id1169736662938\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169736595961\">Find the derivative of [latex]h(x)=(2x^3+2x-1)^4[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739325717\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739325717\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739325717\">[latex]h^{\\prime}(x)=4(2x^3+2x-1)^3(6x^{2}+2)=8(3x^{2}+1)(2x^3+2x-1)^3[\/latex]<\/p>\n<div id=\"fs-id1169739274677\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1169739179049\">Use the previous example with [latex]g(x)=2x^3+2x-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/CE8oqftYNvQ?controls=0&amp;start=136&amp;end=185&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">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\/3.6TheChainRule136to185_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>.<\/div>\n<\/div>\n<div id=\"fs-id1169739302258\" class=\"textbook exercises\">\n<h3>Example: Using the Chain and Power Rules with a Trigonometric Function<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739182335\">Show Solution<\/span><\/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\">[latex]h^{\\prime}(x)=3(\\sin x)^2 \\cos x=3 \\sin^2 x \\cos x[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739190027\" class=\"textbook exercises\">\n<h3>Example: Finding the Equation of a Tangent Line<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739336066\">Show Solution<\/span><\/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>&nbsp;<\/p>\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<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Equation of a Tangent Line.<\/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\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">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>.<\/div>\n<\/div>\n<div id=\"fs-id1169739301868\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739303885\">Find the equation of the line tangent to the graph of [latex]f(x)=(x^2-2)^3[\/latex] at [latex]x=-2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738869705\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738869705\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738869705\">[latex]y=-48x-88[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1169739252012\">Use the preceding example as a guide.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm205911\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=205911&theme=oea&iframe_resize_id=ohm205911&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>The Chain and Trigonometric Functions Combined<\/h2>\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<div id=\"fs-id1169736656617\" class=\"textbook exercises\">\n<h3>Example: Using the Chain Rule on a General Cosine Function<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736657088\">Show Solution<\/span><\/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<div><\/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<\/div>\n<p id=\"fs-id1169739301534\">In the following example we apply the rule that we have just derived.<\/p>\n<div id=\"fs-id1169739301537\" class=\"textbook exercises\">\n<h3>Example: Using the Chain Rule on a Cosine Function<\/h3>\n<p id=\"fs-id1169739301547\">Find the derivative of [latex]h(x)= \\cos (5x^2)[\/latex].<\/p>\n<div id=\"fs-id1169739111373\" class=\"equation unnumbered\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q847788\">Show Solution<\/span><\/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].<br \/>\nUsing the result from the previous example, [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<\/div>\n<div id=\"fs-id1169739333921\" class=\"textbook exercises\">\n<h3>Example: Using the Chain Rule on Another Trigonometric Function<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739300092\">Show Solution<\/span><\/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\">[latex]h^{\\prime}(x)= \\sec (g(x)) \\tan (g(x))g^{\\prime}(x)[\/latex].<\/div>\n<p>&nbsp;<\/p>\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\">[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<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using the Chain Rule on Another Trigonometric Function.<\/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\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" class=\"hidden-answer\" style=\"display: none\">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>.<\/div>\n<\/div>\n<div id=\"fs-id1169739188144\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739188151\">Find the derivative of [latex]h(x)= \\sin (7x+2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q166577\">Hint<\/span><\/p>\n<div id=\"q166577\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736607578\">Apply the chain rule to [latex]h(x)= \\sin g(x)[\/latex] first and then use [latex]g(x)=7x+2[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q232193\">Show Solution<\/span><\/p>\n<div id=\"q232193\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]h^{\\prime}(x)=7 \\cos (7x+2)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169736610159\">At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in the last three examples. For convenience, formulas are also given in Leibniz\u2019s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz\u2019s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.<\/p>\n<div id=\"fs-id1169736610174\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Using the Chain Rule with Trigonometric Functions<\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<h2>The Chain and Product Rules Combined<\/h2>\n<div id=\"fs-id1169739298047\" class=\"textbook exercises\">\n<h3>Example: Combining the Chain Rule with the Product Rule<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739345782\">Show Solution<\/span><\/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>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Combining the Chain Rule with the Product Rule.<\/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\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266836\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266836\" class=\"hidden-answer\" style=\"display: none\">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>.<\/div>\n<\/div>\n<div id=\"fs-id1169736603462\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169736603469\">Find the derivative of [latex]h(x)=\\dfrac{x}{(2x+3)^3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3778210\">Hint<\/span><\/p>\n<div id=\"q3778210\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736603573\">Start out by applying the quotient rule. Remember to use the chain rule to differentiate the denominator.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736603515\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736603515\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736603515\">[latex]h^{\\prime}(x)=\\frac{3-4x}{(2x+3)^4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm206007\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206007&theme=oea&iframe_resize_id=ohm206007&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Applying the Chain Rule Multiple Times<\/h2>\n<p id=\"fs-id1169739304867\">We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple times.<\/p>\n<p id=\"fs-id1169739304874\">In general terms, first we let<\/p>\n<div id=\"fs-id1169739304877\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k(x)=h(f(g(x)))[\/latex]<\/div>\n<p>Then, applying the chain rule once we obtain<\/p>\n<div id=\"fs-id1169739304927\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=\\frac{d}{dx}(h(f(g(x))))=h^{\\prime}(f(g(x))) \\cdot \\frac{d}{dx}(f(g(x)))[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736597613\">Applying the chain rule again, we obtain<\/p>\n<div id=\"fs-id1169736597616\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=h^{\\prime}(f(g(x)))f^{\\prime}(g(x))g^{\\prime}(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div id=\"fs-id1169736597703\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Chain Rule for a Composition of Three Functions<\/h3>\n<hr \/>\n<p id=\"fs-id1169736597709\">For all values of [latex]x[\/latex] for which the function is differentiable, if<\/p>\n<div id=\"fs-id1169736597716\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k(x)=h(f(g(x)))[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1169736597763\">then<\/p>\n<div id=\"fs-id1169736597766\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=h^{\\prime}(f(g(x)))f^{\\prime}(g(x))g^{\\prime}(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1169736658568\">In other words, we are applying the chain rule twice.<\/p>\n<\/div>\n<p id=\"fs-id1169736658572\">Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, <em>remember, we always work from the outside in, taking one derivative at a time.<\/em><\/p>\n<div id=\"fs-id1169736658580\" class=\"textbook exercises\">\n<h3>Example: Differentiating a Composite of Three Functions<\/h3>\n<p id=\"fs-id1169736658589\">Find the derivative of [latex]k(x)=\\cos^4 (7x^2+1)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736658640\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736658640\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736658640\">First, rewrite [latex]k(x)[\/latex] as<\/p>\n<div id=\"fs-id1169736658656\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k(x)=(\\cos(7x^2+1))^4[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739275169\">Then apply the chain rule several times.<\/p>\n<div id=\"fs-id1169739275173\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}k^{\\prime}(x) & =4(\\cos(7x^2+1))^3(\\frac{d}{dx}\\cos(7x^2+1)) & & & \\text{Apply the chain rule.} \\\\ & =4(\\cos(7x^2+1))^3(\u2212\\sin(7x^2+1))(\\frac{d}{dx}(7x^2+1)) & & & \\text{Apply the chain rule.} \\\\ & =4(\\cos(7x^2+1))^3(\u2212\\sin(7x^2+1))(14x) & & & \\text{Apply the chain rule.} \\\\ & =-56x \\sin(7x^2+1) \\cos^3 (7x^2+1) & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>Don&#8217;t forget that\u00a0[latex]{\\cos }^{n}t[\/latex] is a commonly used shorthand notation for [latex]{\\left(\\cos \\left(t\\right)\\right)}^{n}[\/latex]. When we write it without the shorthand notation, we can clearly see why the chain rule is necessary in these situations.<\/div>\n<div><\/div>\n<div><\/div>\n<div id=\"fs-id1169739350744\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739350751\">Find the derivative of [latex]h(x)=\\sin^6 (x^3)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9044612\">Hint<\/span><\/p>\n<div id=\"q9044612\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739350858\">Rewrite [latex]h(x)=\\sin^6 (x^3)=(\\sin(x^3))^6[\/latex] and use the previous example as a guide.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739350792\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739350792\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739350792\">[latex]h^{\\prime}(x)=18x^2 \\sin^5 (x^3) \\cos (x^3)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm206008\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206008&theme=oea&iframe_resize_id=ohm206008&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1169736593542\" class=\"textbook exercises\">\n<h3>Example: Using the Chain Rule in a Velocity Problem<\/h3>\n<p id=\"fs-id1169736593552\">A particle moves along a coordinate axis. Its position at time [latex]t[\/latex] is given by [latex]s(t)= \\sin (2t)+ \\cos (3t)[\/latex]. What is the velocity of the particle at time [latex]t=\\frac{\\pi}{6}[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736593621\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736593621\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736593621\">To find [latex]v(t)[\/latex], the velocity of the particle at time [latex]t[\/latex], we must differentiate [latex]s(t)[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169736593661\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=s^{\\prime}(t)=2 \\cos(2t)-3 \\sin(3t)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739266652\">Substituting [latex]t=\\frac{\\pi}{6}[\/latex] into [latex]v(t)[\/latex], we obtain [latex]v(\\frac{\\pi}{6})=-2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using the Chain Rule in a Velocity Problem.<\/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=725&amp;end=835&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266837\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266837\" class=\"hidden-answer\" style=\"display: none\">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\/3.6TheChainRule725to835_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>.<\/div>\n<\/div>\n<div id=\"fs-id1169739266711\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739266719\">A particle moves along a coordinate axis. Its position at time [latex]t[\/latex] is given by [latex]s(t)= \\sin (4t)[\/latex]. Find its acceleration at time [latex]t[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3388209\">Hint<\/span><\/p>\n<div id=\"q3388209\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736655167\">Acceleration is the second derivative of position.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739266765\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739266765\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739266765\">[latex]a(t)=-16 \\sin (4t)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Proof of the Chain Rule<\/h3>\n<p id=\"fs-id1169736655179\">At this point, we present a very informal proof of the chain rule. For simplicity\u2019s sake we ignore certain issues: For example, we assume that [latex]g(x)\\ne g(a)[\/latex] for [latex]x\\ne a[\/latex] in some open interval containing [latex]a[\/latex]. We begin by applying the limit definition of the derivative to the function [latex]h(x)[\/latex] to obtain [latex]h^{\\prime}(a)[\/latex]:<\/p>\n<div id=\"fs-id1169736655258\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{x-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739305452\">Rewriting, we obtain<\/p>\n<div id=\"fs-id1169739305455\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)} \\cdot \\dfrac{g(x)-g(a)}{x-a}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739305587\">Although it is clear that<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{g(x)-g(a)}{x-a}=g^{\\prime}(a)[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736662280\">it is not obvious that<\/p>\n<div id=\"fs-id1169736662283\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)}=f^{\\prime}(g(a))[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736662391\">To see that this is true, first recall that since [latex]g[\/latex] is differentiable at [latex]a, \\, g[\/latex] is also continuous at [latex]a[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169736662414\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}g(x)=g(a)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739303393\">Next, make the substitution [latex]y=g(x)[\/latex] and [latex]b=g(a)[\/latex] and use change of variables in the limit to obtain<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)}=\\underset{y\\to b}{\\lim}\\dfrac{f(y)-f(b)}{y-b}=f^{\\prime}(b)=f^{\\prime}(g(a))[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739335832\">Finally,<\/p>\n<div id=\"fs-id1169739335835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(a)=\\underset{x\\to a}{\\lim}\\dfrac{f(g(x))-f(g(a))}{g(x)-g(a)} \\cdot \\dfrac{g(x)-g(a)}{x-a}=f^{\\prime}(g(a))g^{\\prime}(a)[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<div><\/div>\n<div id=\"fs-id1169736613849\" class=\"textbook exercises\">\n<h3>Example: Using the Chain Rule with Functional Values<\/h3>\n<p id=\"fs-id1169736613859\">Let [latex]h(x)=f(g(x))[\/latex]. If [latex]g(1)=4, \\, g^{\\prime}(1)=3[\/latex], and [latex]f^{\\prime}(4)=7[\/latex], find [latex]h^{\\prime}(1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736613978\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736613978\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736613978\">Use the chain rule, then substitute.<\/p>\n<div id=\"fs-id1169736613982\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}h^{\\prime}(1) & =f^{\\prime}(g(1))g^{\\prime}(1) & & & \\text{Apply the chain rule.} \\\\ & =f^{\\prime}(4) \\cdot 3 & & & \\text{Substitute} \\, g(1)=4 \\, \\text{and} \\, g^{\\prime}(1)=3. \\\\ & =7 \\cdot 3 & & & \\text{Substitute} \\, f^{\\prime}(4)=7. \\\\ & =21 & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using the Chain Rule with Functional Values.<\/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=842&amp;end=920&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266838\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266838\" class=\"hidden-answer\" style=\"display: none\">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\/3.6TheChainRule842to920_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>.<\/div>\n<\/div>\n<div id=\"fs-id1169739341441\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739341448\">Let [latex]h(x)=f(g(x))[\/latex]. If [latex]g(2)=-3, \\, g^{\\prime}(2)=4[\/latex], and [latex]f^{\\prime}(-3)=7[\/latex], find [latex]h^{\\prime}(2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739353312\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739353312\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739353312\">28<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1169739353326\">Follow the example of using the chain rule in a velocity problem.<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-349\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.6 The Chain Rule. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":26,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.6 The Chain Rule\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-349","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/349","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":27,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/349\/revisions"}],"predecessor-version":[{"id":4814,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/349\/revisions\/4814"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/35"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/349\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=349"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=349"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=349"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=349"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}