{"id":37821,"date":"2018-02-13T21:20:50","date_gmt":"2018-02-13T21:20:50","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=37821"},"modified":"2026-03-26T09:45:42","modified_gmt":"2026-03-26T14:45:42","slug":"implicit-differentiation","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/implicit-differentiation\/","title":{"rendered":"Implicit Differentiation"},"content":{"rendered":"<style>\nmjx-mfrac {\n  font-size: 1.2em;\n}\n<\/style>\n\n\t\t\t<div id=\"mmDeferVideoEncompass_PXnBdiMZ2wM\" 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_PXnBdiMZ2wM\" data-source-videoID=\"PXnBdiMZ2wM\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Implicit Differentiation Video\" height=\"720\" width=\"1280\" class=\"size-full\" data-matomo-title = \"Implicit Differentiation\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_PXnBdiMZ2wM:hover {cursor:pointer;} img#videoThumbnailImage_PXnBdiMZ2wM {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/impdif-thumb.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_PXnBdiMZ2wM\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_PXnBdiMZ2wM\" 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\",\"Implicit Differentiation\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_PXnBdiMZ2wM\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_PXnBdiMZ2wM\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_PXnBdiMZ2wM\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction fI7_Function() {\n  var x = document.getElementById(\"fI7\");\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=\"fI7_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=\"fI7\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Implicit_Differentiation_Example_1\" class=\"smooth-scroll\">Implicit Differentiation Example\u00a01<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Implicit_Differentiation_Example_2\" class=\"smooth-scroll\">Implicit Differentiation Example\u00a02<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Implicit_Differentiation_Example_3\" class=\"smooth-scroll\">Implicit Differentiation Example\u00a03<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Review\" class=\"smooth-scroll\">Review<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Implicit_Differentiation_Practice_Questions\" class=\"smooth-scroll\">Implicit Differentiation Practice Questions<\/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<p>If you&#8217;ve watched our other videos on derivatives, you should have a fair understanding of how to find derivatives of functions. <\/p>\n<p>To review, these functions are written as \\(f(x)\\) and you\u2019ll find \\(f'(x)\\). Equivalently, the function may be called \\(y\\), and you\u2019ll find \\(y&#8217;\\), where \\(f(x)\\) or \\(y\\) is written in terms of \\(x\\).<\/p>\n<p>Sometimes, though, you may need to take the derivative of a function that is not in this format. For example, the relation \\(2xy-y^{2}=5x\\) features multiple terms with \\(y\u2019s\\), and it cannot simply be rearranged to give \\(y\\) as a function of \\(x\\). For these, we have to use a process called \u201cimplicit differentiation\u201d to find the derivative of \\(y\\) with respect to \\(x\\), or \\(\\frac{dy}{dx}\\).<\/p>\n<p>Today we\u2019ll be building on some of the concepts we\u2019ve covered in other videos, including the chain rule and the product rule.<\/p>\n<h2><span id=\"Implicit_Differentiation_Example_1\" class=\"m-toc-anchor\"><\/span>Implicit Differentiation Example&nbsp;1<\/h2>\n<h3><span id=\"Step_1\" class=\"m-toc-anchor\"><\/span>Step 1<\/h3>\n<p>\nLet\u2019s start by considering the relation \\(6xy-x^{2}=3y\\) and finding \\(\\frac{dy}{dx}\\) through the process of implicit differentiation.<\/p>\n<p>To do this, we will want to take the derivative of each term in the relation. We can start by writing \\(\\frac{d}{dx}\\) in front of every single term. Remember, \\(\\frac{d}{dx}\\) is just a notation for saying \u201ctake the derivative with respect to \\(x\\).\u201d<\/p>\n<p>When we do this, we get \\(\\frac{d}{dx}6xy-\\frac{d}{dx}x^{2}=\\frac{d}{dx}3y\\).<\/p>\n<h3><span id=\"Step_2\" class=\"m-toc-anchor\"><\/span>Step 2<\/h3>\n<p>\nThe second step in implicit differentiation is taking each of these derivatives.<\/p>\n<p>Let\u2019s take a look at these terms one at a time. <\/p>\n<h4><span id=\"Term_1\" class=\"m-toc-anchor\"><\/span>Term 1<\/h4>\n<p>\nFirst, we have the derivative of \\(6xy\\) with respect to \\(x\\), but how can we take the derivative with respect to \\(x\\) if there&#8217;s a \\(y\\) in the term? To do this, let\u2019s remember that when we talk about \\(y\\), we typically are talking about \\(y\\) itself being a function of \\(x\\). With implicit differentiation, the same idea applies. We are going to consider all \\(y\\)\u2019s to be implicit functions of \\(x\\). <\/p>\n<p>For this term then, we have \\(6x\\) times a function of \\(x\\). As with derivatives of other products, this calls for the product rule. Remember, the product rules says that for two functions multiplied together, the derivative is the first times the derivative of the second, plus the second times the derivative of the first. So, \u201cthe first times the derivative of the second\u201d gives us \\(6x\\) times \\(\\frac{dy}{dx}\\), because the derivative of \\(y\\), \\(\\frac{dd}{xy}\\), can be written as \\(\\frac{dy}{dx}\\). <\/p>\n<p>Continuing with the product rule, we add \u201cthe second times the derivative of the first,\u201d which gives us \\(y \\times 6\\). Putting these together, we have \\(6x\\frac{dy}{dx}+6y\\) as the derivative of the first term.<\/p>\n<h4><span id=\"Term_2\" class=\"m-toc-anchor\"><\/span>Term 2<\/h4>\n<p>\nNext, we have \\(-\\frac{d}{dx}x^{2}\\). Since this term doesn\u2019t have any \\(y\\)\u2019s, we can just take its derivative normally and get \\(-2x\\).<\/p>\n<h4><span id=\"Term_3\" class=\"m-toc-anchor\"><\/span>Term 3<\/h4>\n<p>\nFinally, on the right side, we have the derivative \\(\\frac{d}{dx}3y\\). Remember, since the derivative of \\(y\\) is \\(\\frac{dy}{dx}\\), we get \\(3\\frac{dy}{dx}\\).<\/p>\n<p>Putting all of the pieces together, we now have \\(6x\\frac{dy}{dx}+6y-2x=3\\frac{dy}{dx}\\). <\/p>\n<h3><span id=\"Step_3\" class=\"m-toc-anchor\"><\/span>Step 3<\/h3>\n<p>\nSince the goal is finding \\(\\frac{dy}{dx}\\), we now just need to isolate \\(\\frac{dy}{dx}\\) on one side of the equation.<\/p>\n<p>Let\u2019s move \\(6x\\frac{dy}{dx}\\) to the other side of the equation so we can have all the \\(\\frac{dy}{dx}\\) together. To do this, we subtract \\(6x\\frac{dy}{dx}\\) from both sides of the equation.<\/p>\n<div class=\"examplesentence\">\\(6y-2x=3\\frac{dy}{dx}-6x\\frac{dy}{dx}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFrom here, we can factor out the \\(\\frac{dy}{dx}\\).<\/p>\n<div class=\"examplesentence\">\\(6y-2x=\\frac{dy}{dx}(3-6x)\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, divide both sides by \\(3-6x\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{6y-2x}{3-6x}=\\frac{dy}{dx}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd just like that, we have the solution! <\/p>\n<p>The three steps of implicit differentiation are writing \\(\\frac{d}{dx}\\) in front of each term, taking the derivative of each term with respect to \\(x\\), and then solving for \\(\\frac{dy}{dx}\\).<\/p>\n<h2><span id=\"Implicit_Differentiation_Example_2\" class=\"m-toc-anchor\"><\/span>Implicit Differentiation Example&nbsp;2<\/h2>\n<p>\nLet\u2019s try this with another example.<\/p>\n<p>Use implicit differentiation to find \\(\\frac{dy}{dx}\\) for the relation \\(y^3-4x=xy\\).<\/p>\n<p>Just like in the last example, we can start the implicit differentiation process by writing \\(\\frac{d}{dx}\\) in front of each term. This gives us:<\/p>\n<div class=\"examplesentence\">\\(\\frac{d}{dx}y^3-\\frac{d}{dx}4x=\\frac{d}{dx}xy\\)<\/div>\n<p>\n&nbsp;<br \/>\nNext, we need to take the derivative of each term with respect to \\(x\\). <\/p>\n<p>Notice that the first term is \\(y^3\\). Unlike \\(y^1\\), we cannot simply take the derivative to be \\(\\frac{dy}{dx}\\). Since we are taking the derivative with respect to \\(x\\), we have to be careful and remember that \\(y\\) is a function of \\(x\\). With this first term, we are essentially seeing a function, \\(y\\), inside another function, a cube. Sound familiar?<\/p>\n<h3><span id=\"Using_the_Chain_Rule\" class=\"m-toc-anchor\"><\/span>Using the Chain Rule<\/h3>\n<p>\nBecause we have a function inside a function, we need to use the chain rule for this term!<\/p>\n<p>Remember, the chain rule says that the derivative of a composite function, \\(f(g(x))\\), is equal to the derivative of the outside function with the inside function unchanged, times the derivative of the inside function.<\/p>\n<p>So in this problem, the derivative of \\(y^3\\) with respect to \\(x\\) is equal to the derivative of the outer cube function, \\(3y^2\\), times the derivative of the inner function with respect to \\(x\\).<\/p>\n<p>This is a very important step to remember with implicit differentiation since it comes up quite frequently. In fact, any time you have to take the derivative of a term involving \\(y\\) to some power other than 1, you\u2019ll need to use the chain rule and multiply by \\(\\frac{dy}{dx}\\).<\/p>\n<p>Let\u2019s take the derivatives of the other terms now. The derivative of \\(-4x\\) with respect to \\(x\\) is \\(-4\\). On the right side, the derivative of \\(xy\\) with respect to \\(x\\) can be found using the product rule.<\/p>\n<p>\u201cThe first times the derivative of the second\u201d gives us \\(x\\frac{dy}{dx}\\), and \u201cthe second times the derivative of the first\u201d is \\(y1\\). So we get \\(x\\cdot \\frac{dy}{dx}+y\\):<\/p>\n<div class=\"examplesentence\">\\(3y^2\\frac{dy}{dx}-4=x\\frac{dy}{dx}+y\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe final step of implicit differentiation is solving this equation for \\(\\frac{dy}{dx}\\). To do this, let\u2019s move \\(x\\frac{dy}{dx}\\) to the left side and move the -4 to the right side. By subtracting \\(x\\frac{dy}{dx}\\) from both sides and adding 4 to both sides, we get this:<\/p>\n<div class=\"examplesentence\">\\(3y^2\\frac{dy}{dx}-x\\frac{dy}{dx}=y+4\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, let\u2019s factor out \\(\\frac{dy}{dx}\\) from both terms on the left side.<\/p>\n<div class=\"examplesentence\">\\(\\frac{dy}{dx}(3y^2-x)=y+4\\)<\/div>\n<p>\n&nbsp;<br \/>\nFinally, we can get \\(\\frac{dy}{dx}\\) by itself by dividing both sides by \\(3y^2-x\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{dy}{dx}=\\frac{y+4}{3y^2-x}\\)<\/div>\n<p>\n&nbsp;<br \/>\nYou may have noticed that both of the solutions in this lesson so far have included the letter \\(y\\) in the derivative. This is nothing to be worried about, though, and is just a consequence of how implicit differentiation works. Rest assured, as long as you follow the three steps properly and remember the chain rule where applicable, you\u2019ll be on the right track.<\/p>\n<h2><span id=\"Implicit_Differentiation_Example_3\" class=\"m-toc-anchor\"><\/span>Implicit Differentiation Example&nbsp;3<\/h2>\n<p>\nLet\u2019s work through one more example. Use implicit differentiation to find \\(\\frac{dy}{dx}\\) for the relation \\(\\text{sin}(y^2)=5x^2+y\\).<\/p>\n<p>First, we can start by writing \\(\\frac{d}{dx}\\) in front of each term.<\/p>\n<div class=\"examplesentence\">\\(\\frac{d}{dx}\\text{sin}(y^2)=\\frac{d}{dx}5&#215;2+\\frac{d}{dx}y\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe first term is going to be the trickiest. Since we have a function, \\(y^2\\), inside the sine function, we are going to need to use the chain rule. The derivative of the outside will be cosine, so we are going to have \\(cos(^y2)\\) times the derivative of the inside. <\/p>\n<div class=\"examplesentence\">\\(\\text{cos}(y^2)\\cdot (\\frac{d}{dx}y^2)\\)<\/div>\n<p>\n&nbsp;<br \/>\nBut what is the derivative of \\(y^2\\) with respect to \\(x\\)? Since \\(y\\) is not to the first power, we are actually going to need to use the chain rule a second time. Now, the outer function is the square, while the inner function is \\(y\\). So the derivative of \\(y^2\\) with respect to \\(x\\) will be \\(2y\\frac{dy}{dx}\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{d}{dx}y^2=2y\\frac{dy}{dx}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe can now rearrange this a little bit on the left side and we\u2019ll get \\(2y \\hspace{2 pt}\\text{cos}(y^2)\\frac{dy}{dx}\\) as the derivative of the first term.<\/p>\n<p>Fortunately, the other two derivatives in this problem are much simpler. The derivative \\(5x^2\\) is \\(10x\\), and the derivative \\(y\\) can be written as \\(\\frac{dy}{dx}\\):<\/p>\n<div class=\"examplesentence\">\\(2y \\hspace{2 pt}\\text{cos}(y^2)\\frac{dy}{dx}=10x+\\frac{dy}{dx}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe final step is, once again, to solve for \\(\\frac{dy}{dx}\\). To do this, let\u2019s subtract \\(\\frac{dy}{dx}\\) from the right side to the left side by subtracting it from both sides.<\/p>\n<div class=\"examplesentence\">\\(2y \\hspace{2 pt}\\text{cos}(y^2)\\frac{dy}{dx}-\\frac{dy}{dx}=10x\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we can factor out \\(\\frac{dy}{dx}\\) from the terms on the left side.<\/p>\n<div class=\"examplesentence\">\\(\\frac{dy}{dx}(2y \\hspace{2 pt}\\text{cos}(y^2)-1)=10x\\)<\/div>\n<p>\n&nbsp;<br \/>\nDividing both sides by \\(2y \\hspace{2 pt}\\text{cos}(y^2)-1\\), we get the solution:<\/p>\n<div class=\"examplesentence\">\\(\\frac{dy}{dx}=\\frac{10x}2y \\hspace{2 pt}\\text{cos}(y^2)-1\\)<\/div>\n<p>\n&nbsp;<\/p>\n<hr>\n<h2><span id=\"Review\" class=\"m-toc-anchor\"><\/span>Review<\/h2>\n<p>\nImplicit differentiation can seem like a weird and tricky topic in calculus, but keep in mind the three main steps. <\/p>\n<ol>\n<li style=\"margin-bottom: 10px;\">Write \\(\\frac{d}{dx}\\) in front of all terms in the relation.<\/li>\n<li style=\"margin-bottom: 10px;\">Take the derivative of each term, taking special care to remember that \\(y\\) is a function of \\(x\\) and will often require the chain rule for proper execution.\n<li>Solve the resulting equation for \\(\\frac{dy}{dx}\\).<\/li>\n<\/ol>\n<p>As always, with a little practice, you should feel more and more confident with implicit differentiation.<\/p>\n<p>I hope this video was helpful. Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Implicit_Differentiation_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Implicit Differentiation Practice Questions<\/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 \/>\nAssume that \\(y\\) is a function of \\(x\\). Use implicit differentiation to find \\(\\frac{dy}{dx}\\) given the equation \\(4xy+y=x^2\\). <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-1-1\">\\(\\frac{dy}{dx}=\\frac{2x-4y}{4x+1}\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\(\\frac{dy}{dx} =2x-4y\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\(\\frac{dy}{dx}=2x-xy\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(\\frac{dy}{dx} = \\frac{2x-4}{4x+y}\\)<\/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 first step of implicit differentiation is taking the derivative of all terms on both sides.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}4xy+ \\frac{d}{dx}y=\\frac{d}{dx}x^2\\)<\/p>\n<p>From here, each term can be handled individually. To find the derivative of \\(4xy\\), notice that the product rule is needed. This gives us:<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}4xy=4x\\cdot \\frac{d}{dx}y+y\\cdot \\frac{d}{dx}4x\\)<\/p>\n<p>The term \\(\\frac{d}{dx}y\\) is equivalent to saying \\(\\frac{dy}{dx}\\), and \\(\\frac{d}{dx}4x\\) is simply 4, so the derivative of \\(4xy\\) is:<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}4xy=4x\\frac{dy}{dx}+4y\\)<\/p>\n<p>Moving along to the second term, \\(\\frac{d}{dx}y\\) is again equivalent to \\(\\frac{dy}{dx}\\).<\/p>\n<p>The final term, on the right side, is easy to differentiate. <\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}x^2=2x\\)<\/p>\n<p>Putting all of the terms\u2019 derivatives together:<\/p>\n<p class=\"longmath\" style=\"text-align: center\">\\(\\frac{d}{dx}4xy+\\frac{d}{dx}y=\\frac{d}{dx}x^2 \\implies 4x\\frac{dy}{dx}+4y+\\frac{dy}{dx}=2x\\)<\/p>\n<p>The goal of this problem is finding \\(\\frac{dy}{dx}\\), which can now be solved for by using some algebra.<\/p>\n<p style=\"text-align:center; line-height: 50px;\">\n\\(4x\\cdot\\frac{dy}{dx}+4y+\\frac{dy}{dx}=2x\\)<br \/>\n\\(4x\\cdot\\frac{dy}{dx}+\\frac{dy}{dx}=2x-4y\\)<br \/>\n\\(\\frac{dy}{dx}(4x+1)=2x-4y\\)<br \/>\n\\(\\frac{dy}{dx}=\\frac{2x-4y}{4x+1}\\)<\/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;<\/p>\n<p style=\"line-height: 35px\">Assume that \\(y\\) is a function of \\(x\\). Use implicit differentiation to find \\(\\frac{dy}{dx}\\) given the equation \\(\\sin(y)+x^2y^3=\\frac{\\pi}{4}\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\(\\frac{dy}{dx}=-\\cos(y)+6xy^2\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(\\frac{dy}{dx}=\\frac{6xy^2}{\\cos(y)}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(\\frac{dy}{dx}=\\frac{\\cos(y)-2xy}{3x^2y}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-4\">\\(\\frac{dy}{dx}=\\frac{-2xy^3}{\\cos(y)+3x^2y^2}\\)<\/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>Step one of implicit differentiation is writing in \\(\\frac{d}{dx}\\) for all terms.<\/p>\n<p style=\"text-align:center; line-height: 50px\">\\(\\frac{d}{dx}\\sin(y)+\\frac{d}{dx}x^2y^3\\)\\(\\:=\\frac{d}{dx}(\\frac{\\pi}{4})\\)<\/p>\n<p>It is easy to see that the right side becomes 0, so we can write that down now.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}\\sin(y)+\\frac{d}{dx}x^2y^3=0\\)<\/p>\n<p>From here, work one term at a time through the left side. The derivative of \\(\\sin(y)\\) can be found by changing to \\(\\cos(y)\\) and applying the chain rule to \\(y\\), since it is a function <em>inside<\/em> the sine function.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}\\sin(y)=\\cos(y)\\cdot\\frac{d}{dx}y=\\cos(y)\\frac{dy}{dx}\\)<\/p>\n<p>Now, the derivative of \\(x^2y^3\\) can be found using the product rule and then the chain rule.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}x^2y^3=x^2\\cdot\\frac{d}{dx}y^3+y^3\\cdot\\frac{d}{dx}x^2\\)<\/p>\n<p>The derivative of \\(x^2\\) is simply \\(2x\\). The derivative of \\(y^3\\) can be found by remembering that \\(y\\) is still a function of \\(x\\), so we apply the chain rule to see:<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}y^3=3y^2\\cdot\\frac{d}{dx}y=3y^2\\frac{dy}{dx}\\)<\/p>\n<p>Therefore:<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}x^2y^3=x^2\\cdot3y^2\\frac{dy}{dx}+2xy^3\\)<\/p>\n<p>Putting together the derivatives from each term of the original problem, we see:<\/p>\n<p class=\"longmath\" style=\"text-align:center;\">\\(\\frac{d}{dx}\\sin(y)+\\frac{d}{dx}x^2y^3=\\frac{d}{dx}(\\frac{\\pi}{4}) \\implies \\cos(y)\\frac{dy}{dx}+3x^2y^2\\frac{dy}{dx}+2xy^3=0\\)<\/p>\n<p>All that\u2019s left now is getting \\(\\frac{dy}{dx}\\) by itself. Be careful in these final steps because the algebra can get messy.<\/p>\n<p style=\"text-align: center; line-height: 50px;\">\n\\(\\text{cos}(y)\\cdot\\frac{dy}{dx}+3x^2y^2\\cdot\\frac{dy}{dx}+2xy^3=0\\)<br \/>\n\\(\\frac{dy}{dx}(\\text{cos}(y)+3x^2y^2)=-2xy^3\\)<br \/>\n\\(\\frac{dy}{dx}=\\frac{-2xy^3}{\\text{cos}(y)+3x^2y^2}\\)<\/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 \/>\nAssume that \\(y\\) is a function of \\(x\\). Use implicit differentiation to find \\(\\frac{dy}{dx}\\) given the equation \\(\\ln(xy^2)=4x\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-3-1\">\\(\\frac{dy}{dx}=\\frac{4xy-y}{2x}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(\\frac{dy}{dx}=4-\\frac{1}{xy^2}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(\\frac{dy}{dx}=4xy^2-y^2\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\(\\frac{dy}{dx}=\\frac{2x-2y}{xy}\\)<\/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>Start by writing in \\(\\frac{d}{dx}\\) for all terms.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}\\ln(xy^2)=\\frac{d}{dx}4x\\)<\/p>\n<p>Beginning from the left side, what is the derivative of \\(\\ln(xy^2)\\)? For this term, we need to use the chain rule twice. The first use is because there is a function within the natural log, \\(xy^2\\).<\/p>\n<p style=\"text-align:center\">\\(\\frac{d}{dx}\\ln(xy^2)=\\frac{1}{xy^2}\\cdot\\frac{d}{dx}xy^2\\)<\/p>\n<p>The derivative of a natural log is 1 divided by the argument, multiplied by the argument\u2019s derivative. The argument\u2019s derivative can be found now by first employing the product rule, then, because \\(y\\) is a <em>function<\/em> of \\(x\\), using the chain rule a second time. <\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}xy^2=x\\cdot\\frac{d}{dx}y^2+y^2\\cdot\\frac{d}{dx}x\\)<\/p>\n<p>Again, \\(y\\) itself is a function, so in taking the derivative of \\(y^2\\), it is necessary to drop the power, multiply by 2, and multiply by \\(y\\)\u2019s derivative.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}y^2=2y\\frac{dy}{dx}\\)<\/p>\n<p>Time to plug this back into the above expressions.<\/p>\n<p style=\"text-align:center; line-height: 50px;\">\n\\(\\frac{d}{dx}xy^2=2xy\\frac{dy}{dx}+y^2\\)<br \/>\n\\(\\frac{d}{dx}\\ln(xy^2)=\\frac{1}{xy^2}\\cdot(2xy\\frac{dy}{dx}+y^2)\\)<\/p>\n<p>This is the left side of the equation we were given. The right side, \\(4x\\), can be differentiated on its own very easily. Putting left and right sides together after taking these derivatives, we have:<\/p>\n<p style=\"text-align:center;\">\\(\\frac{1}{xy^2}\\cdot(2xy\\frac{dy}{dx}+y^2)=4\\)<\/p>\n<p>Now, solve for \\(\\frac{dy}{dx}\\).<\/p>\n<p style=\"text-align:center; line-height: 50px;\">\n\\(\\frac{1}{xy^2}\\cdot(2xy\\frac{dy}{dx}+y^2)=4\\)<br \/>\n\\(2xy\\frac{dy}{dx}+y^2=4xy^2\\)<br \/>\n\\(2xy\\frac{dy}{dx}=4xy^2-y^2\\)<br \/>\n\\(\\frac{dy}{dx}=\\frac{4xy^2-y^2}{2xy}\\)<br \/>\n\\(\\frac{dy}{dx}=\\frac{4xy-y}{2x}\\)<\/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 \/>\nAssume that \\(y\\) is a function of \\(x\\). Use implicit differentiation to find \\(\\frac{dy}{dx}\\) given the equation \\(\\frac{3y}{\\sin(y)}=x\\). <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\(\\frac{dy}{dx}=\\frac{\\cos(y)}{3\\cos(y)-y\\sin(y)}\\)<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\(\\frac{dy}{dx}=x\\cos(y)+\\frac{1}{3}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-3\">\\(\\frac{dy}{dx}=\\frac{\\sin^2(y)}{3\\sin(y)-3y\\cos(y)}\\)<\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\(\\frac{dy}{dx}=\\frac{3\\sin(y)-3y\\cos(y)}{\\sin^2(y)}\\)<\/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>First, write \\(\\frac{d}{dx}\\) before each term.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}(\\frac{3y}{\\sin(y)})=\\frac{d}{dx}x\\)<\/p>\n<p>On the left side, we find the derivative by using the quotient rule.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}(\\frac{3y}{\\sin(y)})=\\frac{\\sin(y) \\cdot \\frac{d}{dx}3y-3y\\cdot \\frac{d}{dx}\\sin(y)}{\\sin^2(y)}\\)<\/p>\n<p>The derivative of \\(3y\\) is \\(3\\frac{dy}{dx}\\). The derivative of \\(\\sin(y)\\) is \\(\\cos(y)\\frac{dy}{dx}\\).<\/p>\n<p>Therefore, the derivative simplifies to:<\/p>\n<p style=\"text-align: center;\">\\(\\frac{d}{dx}(\\frac{3y}{\\sin(y)})=\\frac{3\\sin(y)\\frac{dy}{dx}-3y\\cos(y)\\frac{dy}{dx}}{\\sin^2(y)}\\)<\/p>\n<p>On the right side, the derivative of \\(x\\) is simply 1. With this knowledge of both sides\u2019 derivatives:<\/p>\n<p class=\"longmath\" style=\"text-align: center;\">\\(\\frac{d}{dx}(\\frac{3y}{\\sin(y)})=\\frac{d}{dx}x \\implies \\frac{3\\sin(y)\\frac{dy}{dx}-3y\\cos(y)\\frac{dy}{dx}}{\\sin^2(y)}=1\\)<\/p>\n<p>Now, solve for \\(\\frac{dy}{dx}\\) by factoring it out of the numerator, then dividing everything away.<\/p>\n<p style=\"text-align: center; line-height: 50px;\">\n\\(\\frac{\\frac{dy}{dx}(\\sin(y)-3y\\cos(y))}{\\sin^2(y)}=1\\)<br \/>\n\\(\\frac{dy}{dx}(3\\sin(y)-3y\\cos(y))=\\sin^2(y)\\)<br \/>\n\\(\\frac{dy}{dx}=\\frac{\\sin^2(y)}{3\\sin(y)-3y\\cos(y)}\\)<\/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 \/>\nAssume that \\(y\\) is a function of \\(x\\). Use implicit differentiation to find \\(\\frac{dy}{dx}\\) given the equation \\(\\sqrt{x^3y^2}=6x^2\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-5-1\">\\(\\frac{dy}{dx}=\\frac{24\\sqrt{x}-3y}{2x}\\)<\/div><div class=\"PQ\"  id=\"PQ-5-2\">\\(\\frac{dy}{dx}=\\frac{12x-\\sqrt{x}}{2y}\\)<\/div><div class=\"PQ\"  id=\"PQ-5-3\">\\(\\frac{dy}{dx}=12xy-\\frac{3x^2}{2}\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">\\(\\frac{dy}{dx}=\\frac{1}{12}x-6x^2y\\)<\/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>First, write \\(\\frac{d}{dx}\\) in front of each term.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}\\sqrt{x^3y^2}=\\frac{d}{dx}6x^2\\)<\/p>\n<p>To find the derivative of the term on the left-hand side, apply the chain rule.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}\\sqrt{x^3y^2}=\\frac{1}{2\\sqrt{x^3y^2}}\\cdot \\frac{d}{dx}x^3y^2\\)<\/p>\n<p>The derivative of \\(x^3y^2\\) can then be found using the product rule.<\/p>\n<p style=\"text-align:center;\">\\(\\frac{d}{dx}x^3y^2=x^3\\cdot \\frac{d}{dx}y^2+y^2\\cdot \\frac{d}{dx}x^3\\)<\/p>\n<p>The derivative of \\(x^3\\) is of course \\(3x^2\\), and we use implicit differentiation and the chain rule to find that the derivative of \\(y^2\\) is \\(2y\\frac{dy}{dx}\\).<\/p>\n<p>So the derivative of \\(x^3y^2\\) simplifies to:<\/p>\n<p style=\"text-align: center;\">\\(\\frac{d}{dx}x^3y^2=2x^3y\\frac{dy}{dx}+3x^2y^2\\).<\/p>\n<p>Plug this back into the derivative of the first term:<\/p>\n<p class=\"longmath\" style=\"text-align: center;\">\\(\\frac{d}{dx}\\sqrt{x^3y^2}=\\frac{1}{2\\sqrt{x^3y^2}}(2x^3y\\frac{dy}{dx}+3x^2y^2)\\)<\/p>\n<p>Now, find the derivative of the term on the right-hand side of the equation.<\/p>\n<p style=\"text-align: center;\">\\(\\frac{d}{dx}6x^2=12x\\)<\/p>\n<p>Putting both sides together now, we have:<\/p>\n<p style=\"text-align: center;\">\\(\\frac{1}{2\\sqrt{x^3y^2}}(2x^3y\\frac{dy}{dx}+3x^2y^2)=12x\\)<\/p>\n<p>All that\u2019s left to do is solve for \\(\\frac{dy}{dx}\\).<\/p>\n<p style=\"text-align: center; line-height: 60px;\">\n\\(\\frac{1}{2\\sqrt{x^3y^2}}(2x^3y\\frac{dy}{dx}+3x^2y^2)=12x\\)<br \/>\n\\((2x^3y\\frac{dy}{dx}+3x^2y^2)=12x(2\\sqrt{x^3y^2})\\)<br \/>\n\\(2x^3y\\frac{dy}{dx}=24x\\sqrt{x^3y^2}-3x^2y^2\\)<br \/>\n\\(\\frac{dy}{dx}=\\frac{24x\\sqrt{x^3y^2}-3x^2y^2}{2x^3y}\\)<\/p>\n<p>This result can be simplified a bit further by pulling an \\(x\\) and a \\(y\\) from the radical, and cancelling some \\(x\\)\u2019s and \\(y\\)\u2019s in the numerator and denominator.<\/p>\n<p style=\"text-align: center; line-height: 60px;\">\n\\(\\frac{dy}{dx}=\\frac{24x^2y\\sqrt{x}-3x^2y^2}{2x^3y}\\)<br \/>\n\\(\\frac{dy}{dx}=\\frac{24\\sqrt{x}-3y}{2x}\\)<\/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":237598,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-37821","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\/37821","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=37821"}],"version-history":[{"count":7,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/37821\/revisions"}],"predecessor-version":[{"id":287351,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/37821\/revisions\/287351"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/237598"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=37821"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}