{"id":37970,"date":"2018-02-19T20:12:32","date_gmt":"2018-02-19T20:12:32","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=37970"},"modified":"2026-03-28T10:46:47","modified_gmt":"2026-03-28T15:46:47","slug":"chain-rule","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/chain-rule\/","title":{"rendered":"The Chain Rule &#8211; An Integral Part of Calculus"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_U6n9TKMjIfs\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_U6n9TKMjIfs\" data-source-videoID=\"U6n9TKMjIfs\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"The Chain Rule &#8211; An Integral Part of Calculus Video\" height=\"720\" width=\"1280\" class=\"size-full\" data-matomo-title = \"The Chain Rule &#8211; An Integral Part of Calculus\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_U6n9TKMjIfs:hover {cursor:pointer;} img#videoThumbnailImage_U6n9TKMjIfs {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/10\/Chain-Rule-thumb.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_U6n9TKMjIfs\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_U6n9TKMjIfs\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"The Chain Rule &#8211; An Integral Part of Calculus\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_U6n9TKMjIfs\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_U6n9TKMjIfs\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_U6n9TKMjIfs\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction cSB_Function() {\n  var x = document.getElementById(\"cSB\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"cSB_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"cSB\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Chain_Rule\" class=\"smooth-scroll\">Chain Rule<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Composition_of_Functions\" class=\"smooth-scroll\">Composition of Functions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#What_is_the_Chain_Rule\" class=\"smooth-scroll\">What is the Chain Rule?<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Chain_Rule_Examples\" class=\"smooth-scroll\">Chain Rule Examples<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Review\" class=\"smooth-scroll\">Review<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Chain_Rule_Problems\" class=\"smooth-scroll\">Chain Rule Problems<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Chain_Rule\" class=\"m-toc-anchor\"><\/span>Chain Rule<\/h2>\n<p>Up to this point, you should\u2019ve learned how to take the derivatives of various types of functions, including <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/definition-of-the-derivative\/\">polynomials<\/a>, <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/product-and-quotient-rule\/\">products<\/a>, <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/derivatives-of-exponential-and-log-functions\/\">natural logs<\/a>, and more. But how would we go about taking the derivative of something like this, where one function is nested inside of another?<\/p>\n<div class=\"examplesentence\">\\(f(x)=ln(5x^{2}+1)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>In this video, I\u2019m going to show you how to apply the chain rule for problems such as these.<\/p>\n<h2><span id=\"Composition_of_Functions\" class=\"m-toc-anchor\"><\/span>Composition of Functions<\/h2>\n<p>\nBefore we get started, let\u2019s briefly review composition of functions. Composition of functions can be described as nesting one function inside another and is usually denoted with a little \u201co\u201d between the two functions. For example, if \\(f(x)=x^{3}\\text{    }\\)and \\(\\text{  }g(x)=8x-2\\), then we would say:<\/p>\n<div class=\"examplesentence\">\\(f\\circ g=f(g(x))=(8x-2)^{3}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>To compose one function with another, it&#8217;s helpful to write them in nested form like \\(f(g(x))\\). From here, we can replace \\(g(x)\\) with \\(8x-2\\), which gives us \\(f(8x-2)\\), and plug that in everywhere we see an \\(x\\) in the function \\(f(x)\\).<\/p>\n<div class=\"examplesentence\">\\(f\\circ g=f(g(x))=(8x-2)^{3}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>In a similar way, we could write \\(g\\circ f\\) as \\(g\\circ f = g(f(x))=g(x^{3})=8x^{3}-2\\).<\/p>\n<h2><span id=\"What_is_the_Chain_Rule\" class=\"m-toc-anchor\"><\/span>What is the Chain Rule?<\/h2>\n<p>\nThe chain rule is a method which helps us take the derivative of \u201cnested\u201d functions like \\(f(g(x))\\). It states that the derivative of a composite function, like \\(f\\circ g\\), is equal to the derivative of the outer function, composed with the inner function, and this is multiplied by the derivative of the inner function.<\/p>\n<p>Another way of expressing the chain rule is by writing:<\/p>\n<div class=\"examplesentence\">\\((f\\circ g)&#8217;=f'(g(x))\\cdot g'(x)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h2><span id=\"Chain_Rule_Examples\" class=\"m-toc-anchor\"><\/span>Chain Rule Examples<\/h2>\n<h3><span id=\"Chain_Rule_Example_1\" class=\"m-toc-anchor\"><\/span>Chain Rule Example #1<\/h3>\n<p>\nSo, if we wanted to find the derivative of \\(f(g(x))\\), we would first write the derivative of the outer function, the cube, while leaving the inner function alone.<\/p>\n<p>Given that \\(f(x)=x^3\\), then \\(f'(x)=3x^2\\) and \\(f(g(x))&#8217;=3(8x-2)^{2}\\).<\/p>\n<p>Then, multiply this by the derivative of the inner function, \\(8x-2\\), which would be 8.<\/p>\n<p>As a result, \\((f\\circ g)&#8217;=3(8x-2)^2 \\cdot 8\\). This can simplify to \\(24(8x-2)^2\\).<\/p>\n<p>Now, we could have handled this derivative by expanding out \\((8x-2)^{3}\\) and then applying the power rule to the resulting polynomial. However, this expansion would have been time-consuming to calculate. Once you get the hang of the chain rule, you\u2019ll find that it is an excellent time-saver in problems like this.<\/p>\n<h3><span id=\"Chain_Rule_Example_2\" class=\"m-toc-anchor\"><\/span>Chain Rule Example #2<\/h3>\n<p>\nLet\u2019s work through another example. Use the chain rule to find the derivative of the function \\(h(x)=\\sqrt{ln(x)}\\).<\/p>\n<p>First, let\u2019s identify what the outer and inner functions are. Since \\(ln(x)\\) is nested inside the radical, it is the inner function, while the outer function is the square root. The chain rule says that the derivative of a composite function is equal to the derivative of the outer function with the inner function untouched, times the derivative of the inner function. Since the derivative of \\(\\sqrt{x}\\), or \\(x^{\\frac{1}{2}}\\), is \\(\\frac{1}{2}x^{-\\frac{1}{2}}\\), we write the first part of the derivative as:<\/p>\n<div class=\"examplesentence\">\\(f'(g(x))=\\frac{1}{2}(ln(x))^{-\\frac{1}{2}}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Then, multiply by the derivative of the inside. Since the derivative of \\(ln(x)\\) is \\(\\frac{1}{x}\\), we will multiply by \\(\\frac{1}{x}\\).<\/p>\n<div class=\"examplesentence\">\\(f'(g(x))=\\frac{1}{2} (ln(x))^{-\\frac{1}{2}}(\\frac{1}{x})\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Simplifying, we get that \\(f'(g(x))=\\frac{1}{2x}(ln(x))^{-\\frac{1}{2}}\\), or \\(\\frac{1}{2x\\sqrt{ln(x)}}\\).<\/p>\n<h3><span id=\"Chain_Rule_Example_3\" class=\"m-toc-anchor\"><\/span>Chain Rule Example #3<\/h3>\n<p>\nLet\u2019s do another example. Use the chain rule to find the derivative of the function \\(j(x)=e^{8x^{2}}\\). <\/p>\n<p>This function may look a little intimidating, but we have the tools necessary to work through its derivative. In this case, the outer function is \\(f(x)=e^x\\). The inner function is that power: \\(g(x)=8x^{2}\\). <\/p>\n<p>To get the derivative with the chain rule, we need to first write the derivative of the outer function with the inner function untouched. Since the derivative of the outer function, \\(e^{x}\\), is itself, we are just going to write \\(e^{8x^{2}}\\) again:<\/p>\n<div class=\"examplesentence\">\\(j'(x)=e^{8x^{2}} \\cdot g'(x)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Then, we need to multiply this by the derivative of the inner function. Since the derivative of \\(8x^2=16x\\), let\u2019s go ahead and write that down:<\/p>\n<div class=\"examplesentence\">\\(j'(x)=e^{8x^{2}}(16x)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>This simplifies to \\(6xe^{8x^{2}}\\).<\/p>\n<p>Not so bad, right?<\/p>\n<p>Now I want you to try a problem on your own. <\/p>\n<h3><span id=\"Chain_Rule_Example_4\" class=\"m-toc-anchor\"><\/span>Chain Rule Example #4<\/h3>\n<p>\nUsing the chain rule, find the derivative of \\(k(x)=(x^{3}-9)^{11}\\). Remember, you\u2019ll need to handle the outer function\u2019s derivative first, then the inner function\u2019s derivative. Pause the video now to work out the answer yourself, then we will discuss the solution together.<\/p>\n<p>Let\u2019s check your work. What is the outer function? In this problem, it is the eleventh power. The inner function is \\(x^{3}-9\\). The first part of the chain rule dictates that we write the derivative of the outer function with the inner function still inside of it, so we will have \\(11(x^3-9)^{10}\\). Then, we are going to multiply this by the derivative of the inner function, which will be \\(3x^{2}\\).<\/p>\n<div class=\"examplesentence\">\\(k'(x)=11(x^{3}-9)^{10}(3x^{2})\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>If we simplify this a little bit, we\u2019ll get:<\/p>\n<div class=\"examplesentence\">\\(k'(x)=33x^{2}(x^{3}-9)^{10}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>This problem, like the first one in this video, could have been solved by expanding the polynomial to a twelve-term expression and then handling each term individually with the power rule. But I think it\u2019s safe to say that the chain rule is a much better method for finding this derivative.<\/p>\n<p>Let\u2019s work through one last example. <\/p>\n<h3><span id=\"Chain_Rule_Example_5\" class=\"m-toc-anchor\"><\/span>Chain Rule Example #5<\/h3>\n<p>\nLeo is studying population growth in a petri dish of mold spores. The dish starts out with only one spore, but the spore duplicates itself and there is exponential population growth over time. If the number of mold spores on day \\(x\\) can be calculated using the function \\(m(x)=e^{\\sqrt{x}}\\), determine the function \\(m'(x)\\), which represents the growth rate.<\/p>\n<p>Because we are interested in finding a rate function, \\(m'(x)\\), we understand that this is a derivative problem. And since the function \\(m(x)\\) is a composite function, we know that we are going to need the chain rule to find that derivative.<\/p>\n<p>The outer function in this case is \\(f(x)=e^x\\). That power, \\(\\sqrt{x}\\), is the inner function. To find \\(m&#8217;\\), we start by taking the derivative of the outer function while leaving the inner function as is. Once again, the derivative of \\(e^x\\) is itself, so we will start by just writing \\(e^{\\sqrt{x}}\\) again.<\/p>\n<div class=\"examplesentence\">\\(m'(x)=e^{\\sqrt{x}}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Next, we have to multiply this by the derivative of the inner function. Since the derivative of \\(\\sqrt{x}\\) is \\(\\frac{1}{2}x^{-\\frac{1}{2}}\\), the growth rate function \\(m&#8217;\\) is equal to:<\/p>\n<div class=\"examplesentence\">\\(m'(x)=e^{\\sqrt{x}}(\\frac{1}{2}x^{-\\frac{1}{2}})\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>or equivalently:<\/p>\n<div class=\"examplesentence\">\\(m'(x)=\\)<span style=\"font-size: 110%;\">\\(\\frac{e^{\\sqrt{x}}}{2\\sqrt{x}}\\)<\/span><\/div>\n<p>\n&nbsp;<\/p>\n<hr>\n<h2><span id=\"Review\" class=\"m-toc-anchor\"><\/span>Review<\/h2>\n<p>\nRemember, the chain rule helps us take the derivatives of composite (nested) functions. To get these derivatives, first write the derivative of the outer function with the inner function still untouched within it, then multiply this by the derivative of the inner function.<\/p>\n<div class=\"examplesentence\">\\((f\u2218g)&#8217;=f'(g(x))\\cdot g'(x)\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>This technique will come in handy any time you see a function contained in another function, whether under a radical, in a logarithm, or even inside a trig function. As always, a little practice goes a long way, so I encourage you to try some problems on your own.<\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Chain_Rule_Problems\" class=\"m-toc-anchor\"><\/span>Chain Rule Problems<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nUse the chain rule to find the derivative, \\(h&#8217;\\), of the following function: <\/p>\n<div class=\"yellow-math-quote\">\\(h(x)=\\cos(9x^4+17)\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(\\cos (36x^3)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-2\">\\(-36x^3\\sin (9x^4+17)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\(-\\sin (36x^3)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(\\cos (36x^3)-\\sin (9x^4+17)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The function \\(h\\) is the composition of the outer and inner functions:<\/p>\n<ul style=\"list-style-type: none; margin-left: 1.2em\">\n<li style=\"margin-bottom: 8px\">\\(f(x)=\\cos x)\\)<\/li>\n<li>\\(g(x)=9x^4+17\\)<\/li>\n<\/ul>\n<p>The chain rule tells us the following:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(h'(x)=f'(g(x))\\cdot g'(x)\\)\\(\\: = \\text{outer}'(\\text{inner untouched})\\cdot \\text{ inner}&#8217;\\)<\/p>\n<p>Since \\(f'(x)=-\\sin x)\\) and \\(g'(x)=9(4x^3)=36x^3\\), the chain rule gives us:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(h'(x)=-\\sin(9x^4+17) \\cdot (36x^3)\\)\\(\\: =-36x^3\\sin(9x^4+17)\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nUse the chain rule to find the derivative, \\(k&#8217;\\), of the following function:<\/p>\n<div class=\"yellow-math-quote\">\\(k(x)=\\ln (e^x+12)\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\(\\large{\\frac{e^x}{x}}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(\\ln(e^x)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-3\">\\(\\large{\\frac{e^x}{e^x+12}}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\(\\large{\\frac{1}{e^x+12}}\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The function \\(k\\) is the composition of the outer and inner functions:<\/p>\n<ul style=\"list-style-type: none; margin-left: 1.2em\">\n<li style=\"margin-bottom: 8px\">\\(f(x)=\\ln x\\)<\/li>\n<li>\\(g(x)=e^x+12\\)<\/li>\n<\/ul>\n<p>The chain rule tells us the following:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(k'(x)=f'(g(x))\\cdot g'(x)\\)\\(\\: = \\text{outer}'(\\text{inner untouched})\\cdot \\text{ inner}&#8217;\\)<\/p>\n<p>Since \\(f'(x)=\\large{\\frac{1}{x}})\\) and \\(g'(x)=e^x + 0=e^x \\), the chain rule gives us:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(k'(x)= \\dfrac{1}{e^x+12} \\cdot e^x = \\dfrac{e^x}{e^x+12}\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nUse the chain rule to find the derivative, \\(m&#8217;\\), of the following function:<\/p>\n<div class=\"yellow-math-quote\">\\(m(x)=\\sqrt{2x^3-5x^2+13x-92}\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-3-1\">\\(\\large{\\frac{6x^2-10x+13}{2\\sqrt{2x^3-5x^2+13x-92}}}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(\\large{\\frac{1}{2\\sqrt{2x^3-5x^2+13x-92}}}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(\\sqrt{6x^2-10x+13}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\(\\large{\\frac{1}{2\\sqrt{6x^2-10x+13}}}\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The function \\(m\\) is the composition of the outer and inner functions:<\/p>\n<ul style=\"list-style-type: none; margin-left: 1.2em\">\n<li style=\"margin-bottom: 8px\">\\(f(x)=\\sqrt{x}= x^{\\tfrac{1}{2}}\\)<\/li>\n<li>\\(g(x)=2x^3-5x^2+13x-92\\)<\/li>\n<\/ul>\n<p>The chain rule tells us the following:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(m'(x)=f'(g(x))\\cdot g'(x)\\)\\(\\: = \\text{outer}'(\\text{inner untouched})\\cdot \\text{ inner}&#8217;\\)<\/p>\n<p>Since \\(f'(x)=(\\large{\\frac{1}{2}}\\normalsize{)x^{-\\tfrac{1}{2}}}\\) and \\(g'(x)=6x^2 &#8211; 10=x +13 \\), the chain rule gives us:<\/p>\n<p class=\"longmath\" style=\"text-align: center; line-height: 65px\">\\(m'(x)= \\frac{1}{2}(2x^3-5x^2+13x-92)^{-\\tfrac{1}{2}}\\cdot (6x^2-10x+13)\\)\\(\\: =\\dfrac{6x^2-10x+13}{2\\sqrt{2x^3-5x^2+13x-92}}\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA weight suspended from a spring oscillates in such a fashion that its position about the ground at time \\(t\\) seconds is \\(s(t)=20+18\\sin(3t)\\text{ cm}\\). Differentiate the position function to find the velocity of the weight at time \\(t\\) seconds. That is, find \\(v(t)=s'(t)\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\(18\\cos(3t)\\text{ cm\/sec}\\)<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\(18\\cos(3)\\text{ cm\/sec}\\)<\/div><div class=\"PQ\"  id=\"PQ-4-3\">\\(54\\sin(3t)\\text{ cm\/sec}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-4\">\\(54\\cos(3t)\\text{ cm\/sec}\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The function \\(s\\) is the composition of the outer and inner functions:<\/p>\n<ul style=\"list-style-type: none; margin-left: 1.2em\">\n<li style=\"margin-bottom: 8px\">\\(f(t)=20+18\\sin (t)\\)<\/li>\n<li>\\(g(t)=3t\\)<\/li>\n<\/ul>\n<p>The chain rule tells us the following:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(s'(t)=f'(g(t))\\cdot g'(t)\\)\\(\\: = \\text{outer}'(\\text{inner untouched})\\cdot \\text{ inner}&#8217;\\)<\/p>\n<p>Since \\(f'(t)=0+18\\cos (t)=18\\cos (t)\\) and \\(g'(t)=3 \\), the chain rule gives us:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(s'(t)= v(t)=18\\cos (3t) \\cdot 3\\)\\(\\: = 54 \\cos (3t) \\text{ cm\/sec}\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nYou begin a new exercise program to build up your running endurance. You find that after \\(x\\)weeks of this program, the daily distance you can run is \\(d(x)=1.5\\ln (x^2+1)\\text{ miles}\\). What is the derivative \\(d'(x)\\), the function that gives the rate at which your distance is increasing?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">\\(\\large{\\frac{1.5}{x^2+1}}\\normalsize{\\text{ miles\/week}}\\)<\/div><div class=\"PQ\"  id=\"PQ-5-2\">\\(1.5\\ln (2x)\\text{ miles\/week}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-3\">\\(\\large{\\frac{3x}{x^2+1}}\\normalsize{\\text{ miles\/week}}\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">\\(\\large{\\frac{3}{4x}}\\normalsize{\\text{ miles\/week}}\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The function \\(d\\) is the composition of the outer and inner functions:<\/p>\n<ul style=\"list-style-type: none; margin-left: 1.2em\">\n<li style=\"margin-bottom: 8px\">\\(f(x)=1.5\\ln (x)\\)<\/li>\n<li>\\(g(x)=x^2 + 1\\)<\/li>\n<\/ul>\n<p>The chain rule tells us the following:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(d'(x)=f'(g(x))\\cdot g'(x)\\)\\(\\: = \\text{outer}'(\\text{inner untouched})\\cdot \\text{ inner}&#8217;\\)<\/p>\n<p>Since \\(f'(x)=\\large{\\frac{1.5}{x}}\\) and \\(g'(x)=2x \\), the chain rule gives us:<\/p>\n<p class=\"longmath\" style=\"text-align: center\">\\(d'(x)= \\dfrac{1.5}{x^2+1} \\cdot 2x = \\dfrac{3x}{x^2+1} \\text{ miles\/week}\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/calculus\/\">Return to Calculus Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Calculus Videos<\/p>\n","protected":false},"author":1,"featured_media":229047,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-37970","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-calculus-videos","7":"page_category-math-advertising-group","8":"page_category-video-pages-for-study-course-sidebar-ad","9":"page_type-video","10":"content_type-practice-questions","11":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/37970","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=37970"}],"version-history":[{"count":7,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/37970\/revisions"}],"predecessor-version":[{"id":280580,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/37970\/revisions\/280580"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/229047"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=37970"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}