{"id":190,"date":"2019-08-27T18:22:29","date_gmt":"2019-08-27T18:22:29","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=190"},"modified":"2019-12-02T21:20:15","modified_gmt":"2019-12-02T21:20:15","slug":"quotient-rule","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/quotient-rule\/","title":{"rendered":"Quotient Rule"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<title>Quotient Rule for Derivatives &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p>\n<!------------------------>\n\nSuppose we are working with a function $h(x)$ that is a ratio of two\nfunctions $f(x)$ and $g(x)$.\n\n<\/p>\n\n\n\n<center>\n\t<b>How is the derivative of $h(x)$ related to $f(x)$, $g(x)$, and their derivatives?<\/b>\n<\/center>\n\n\n\n<h4 class=\"wp-block-heading\">Quotient Rule<\/h4>\n\n\n\n<p>\n\nLet $f$ and $g$ be differentiable at $x$ with $g(x)\\neq 0$.  Then\n$f\/g$ is differentiable at $x$ and\n\\[\\left[\\frac{f(x)}{g(x)}\\right]&#8217;=\\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}.\\]\n\n<br><\/p>\n\n\n<style>.kt-accordion-id_fb4234-95 .kt-accordion-inner-wrap{column-gap:var(--global-kb-gap-md, 2rem);row-gap:0px;}.kt-accordion-id_fb4234-95 .kt-accordion-panel-inner{border-top-width:0px;border-right-width:0px;border-bottom-width:0px;border-left-width:0px;padding-top:var(--global-kb-spacing-sm, 1.5rem);padding-right:var(--global-kb-spacing-sm, 1.5rem);padding-bottom:var(--global-kb-spacing-sm, 1.5rem);padding-left:var(--global-kb-spacing-sm, 1.5rem);}.kt-accordion-id_fb4234-95 > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > 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.kt-blocks-accordion-header:focus-visible{color:#444444;background:#eeeeee;border-top-color:#eeeeee;border-right-color:#eeeeee;border-bottom-color:#eeeeee;border-left-color:#eeeeee;}.kt-accordion-id_fb4234-95:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle ) .kt-accordion-header-wrap .kt-blocks-accordion-header:hover .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_fb4234-95:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle ) .kt-accordion-header-wrap .kt-blocks-accordion-header:hover .kt-blocks-accordion-icon-trigger:before, body:not(.hide-focus-outline) .kt-accordion-id_fb4234-95:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle ) .kt-blocks-accordion--visible .kt-blocks-accordion-icon-trigger:after, body:not(.hide-focus-outline) 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.kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active{color:#ffffff;background:#444444;border-top-color:#444444;border-right-color:#444444;border-bottom-color:#444444;border-left-color:#444444;}.kt-accordion-id_fb4234-95:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_fb4234-95:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#ffffff;}.kt-accordion-id_fb4234-95:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger{background:#ffffff;}.kt-accordion-id_fb4234-95:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_fb4234-95:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#444444;}@media all and (max-width: 767px){.kt-accordion-id_fb4234-95 .kt-accordion-inner-wrap{display:block;}.kt-accordion-id_fb4234-95 .kt-accordion-inner-wrap .kt-accordion-pane:not(:first-child){margin-top:0px;}}<\/style>\n<div class=\"wp-block-kadence-accordion alignnone\"><div class=\"kt-accordion-wrap kt-accordion-wrap kt-accordion-id_fb4234-95 kt-accordion-has-2-panes kt-active-pane-0 kt-accordion-block kt-pane-header-alignment-left kt-accodion-icon-style-basic kt-accodion-icon-side-right\" style=\"max-width:none\"><div class=\"kt-accordion-inner-wrap\" data-allow-multiple-open=\"true\" data-start-open=\"none\">\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-1 kt-pane_8d284d-76\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">Proof of Quotient Rule<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<p> We will apply the <strong>limit definition of the derivative<\/strong>:<\/p>\n\n\n\n<p style=\"text-align:center\">\n\n  \\[f'(x)=\\lim_{\\Delta x\\to 0}\\, \\frac{f(x+\\Delta x)-f(x)}{\\Delta x}\\]  \n\n<\/p>\n\n\n\n<p> \\begin{eqnarray*} h'(x)=\\left[\\frac{f(x)}{g(x)}\\right]&#8217;&amp;=&amp;\\lim_{h\\to 0}\\, \\frac{\\frac{f(x+h)}{g(x+h)}-\\frac{f(x)}{g(x)}}{h}\\\\ &amp;=&amp; \\lim_{h\\to 0}\\, \\frac{1}{h}\\, \\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\\\ &amp;=&amp; \\lim_{h\\to 0}\\, \\frac{1}{h}\\, \\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\\\  &amp;=&amp; \\lim_{h\\to 0}\\, \\frac{1}{h}\\left[ \\frac{f(x+h)g(x)-f(x)g(x)}{g(x+h)g(x)}-\\frac{f(x)g(x+h)-f(x)g(x)}{g(x+h)g(x)} \\right]\\\\ &amp;=&amp; \\lim_{h\\to 0}\\left[\\frac{1}{g(x+h)}\\right]\\, \\lim_{h\\to 0}\\left[\\frac{f(x+h)-f(x)}{h}\\right]\\\\ &amp;~&amp;\\qquad -\\lim_{h\\to 0} \\left[\\frac{f(x)}{g(x+h)g(x)}\\right]\\, \\lim_{h\\to 0} \\left[\\frac{g(x+h)-g(x)}{h}\\right]\\\\ &amp;=&amp; \\frac{1}{g(x)}\\cdot f'(x)-\\frac{f(x)}{[g(x)]^2}\\cdot g'(x)\\\\ &amp;=&amp;\\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}. \\end{eqnarray*} We have implicitly assumed here that $f'(x)$ and $g'(x)$ exist and that $g(x)\\neq 0$:<\/p>\n\n\n\n<p> \\begin{eqnarray*} h'(x)=\\left[\\frac{f(x)}{g(x)}\\right]&#8217;&amp;=&amp;\\lim_{h\\to 0}\\, \\frac{\\frac{f(x+h)}{g(x+h)}-\\frac{f(x)}{g(x)}}{h}\\\\ &amp;=&amp; \\lim_{h\\to 0}\\, \\frac{1}{h}\\, \\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\\\ &amp;=&amp; \\lim_{h\\to 0}\\, \\frac{1}{h}\\, \\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\\\  &amp;=&amp; \\lim_{h\\to 0}\\, \\frac{1}{h}\\left[ \\frac{f(x+h)g(x)-f(x)g(x)}{g(x+h)g(x)}-\\frac{f(x)g(x+h)-f(x)g(x)}{g(x+h)g(x)} \\right]\\\\ &amp;=&amp; \\lim_{h\\to 0}\\left[\\frac{1}{g(x+h)}\\right]\\, \\lim_{h\\to 0}\\left[\\frac{f(x+h)-f(x)}{h}\\right]\\\\ &amp;~&amp;\\qquad -\\lim_{h\\to 0} \\left[\\frac{f(x)}{g(x+h)g(x)}\\right]\\, \\lim_{h\\to 0} \\left[\\frac{g(x+h)-g(x)}{h}\\right]\\\\ &amp;=&amp; \\frac{1}{g(x)}\\cdot f'(x)-\\frac{f(x)}{[g(x)]^2}\\cdot g'(x)\\\\ &amp;=&amp;\\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}. \\end{eqnarray*} We have implicitly assumed here that $f'(x)$ and $g'(x)$ exist and that $g(x)\\neq 0$.   <\/p>\n<\/div><\/div><\/div>\n<\/div><\/div><\/div>\n\n\n\n<h6 class=\"wp-block-heading\">Examples<\/h6>\n\n\n\n<ul class=\"wp-block-list\"><li> If $\\displaystyle f(x)=\\frac{2x+1}{x-3}$, then \n\t\t\\begin{eqnarray*}\n\t\t\tf'(x)&amp;=&amp;\\frac{(x-3)\\frac{d}{dx}[2x+1]-(2x+1)\\frac{d}{dx}[x-3]}{[x-3]^2}\\\\\n\t\t\t&amp;=&amp; \\frac{(x-3)(2)-(2x+1)(1)}{(x-3)^2}\\\\\n\t\t\t&amp;=&amp; -\\frac{7}{(x-3)^2}.\n\t\t\\end{eqnarray*}\n\n\t<br><br>\n\t<\/li><li> If $\\displaystyle f(x)=\\tan x=\\frac{\\sin x}{\\cos x}$, then \n\t\t\\begin{eqnarray*}\n\t\t\tf'(x)&amp;=&amp;\\frac{\\cos (x)\\frac{d}{dx}[\\sin (x)]-\\sin (x)\\frac{d}{dx} \n\t\t\t[\\cos x]}{[\\cos x ]^2}\\\\\n\t\t\t&amp;=&amp; \\frac{\\cos^2 (x)+\\sin^2 (x)}{\\cos^2 (x)}\\\\\n\t\t\t&amp;=&amp; \\frac{1}{\\cos^2 (x)}\\\\\n\t\t\t&amp;=&amp; \\sec^2 (x),\n\t\t\\end{eqnarray*}\n\t\tverifying the familiar differentiation formula for $\\tan (x)$.\n\n\t<br><br>\n\t<\/li><li> If $\\displaystyle f(x)=\\frac{1}{g(x)}$, then \n\t\t\\begin{eqnarray*}\n\t\t\tf'(x)=\\left[\\frac{1}{g(x)}\\right]&#8217;&amp;=&amp;\\frac{g(x)\\frac{d}{dx}[1]-(1)g'(x)}\n\t\t\t{[g(x)]^2}\\\\\n\t\t\t&amp;=&amp; \\frac{g(x)(0)-(1)g'(x)}{[g(x)]^2}\\\\\n\t\t\t&amp;=&amp; -\\frac{g'(x)}{[g(x)]^2}.\n\t\t\\end{eqnarray*}\n\t\tFor example, $\\displaystyle\n\t\t\\frac{d}{dx}[x^{-4}]=\\frac{d}{dx}\\left[\\frac{1}{x^4}\\right]\n\t\t=-\\frac{\\frac{d}{dx}[x^4]}{[x^4]^2}=-\\frac{4x^3}{x^8}=-\\frac{4}{x^5}=-4x^{-5}$.\n<\/li><\/ul>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<p>\n<\/p><h3>Key Concepts<\/h3>\n<p>\n<\/p><\/center>\n\n\n\n<h4 class=\"wp-block-heading\">Quotient Rule<\/h4>\n\n\n\n<p>\n\nLet $f$ and $g$ be differentiable at $x$ with $g(x) \\neq 0$. Then $f\/g$ is \ndifferentiable at $x$ and\n$$\\left[\\frac{f(x)}{g(x)}\\right]&#8217; = \\frac{g(x)f'(x)-f(x)g'(x)}{\\left[g(x)\\right]^2}.$$\n\n\n<!------------------------>\n\n\n<br>\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ1510\/\">I&#8217;m ready to take the quiz.<\/a>]\n[<a href=\"#top\">I need to review more.<\/a>]<br>\n\n\n<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Quotient Rule for Derivatives &#8211; HMC Calculus Tutorial Suppose we are working with a function $h(x)$ that is a ratio of two functions $f(x)$ and $g(x)$. How is the derivative of $h(x)$ related to $f(x)$, $g(x)$, and their derivatives? Quotient Rule Let $f$ and $g$ be differentiable at $x$ with $g(x)\\neq 0$. Then $f\/g$ is&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":57,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-190","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/190","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=190"}],"version-history":[{"count":6,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/190\/revisions"}],"predecessor-version":[{"id":1095,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/190\/revisions\/1095"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/57"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=190"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=190"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}