{"id":549,"date":"2025-10-09T15:20:56","date_gmt":"2025-10-09T15:20:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=549"},"modified":"2025-11-20T15:12:57","modified_gmt":"2025-11-20T15:12:57","slug":"vector-valued-functions-and-motion-background-youll-need-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-valued-functions-and-motion-background-youll-need-3\/","title":{"raw":"Vector-Valued Functions and Motion: Background You'll Need 3","rendered":"Vector-Valued Functions and Motion: Background You&#8217;ll Need 3"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Apply specific rules to find derivatives of functions multiplied or divided by each other<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 class=\"entry-title\">The Advanced Rules<\/h2>\r\n<h3>The Product Rule<\/h3>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When differentiating a product of two functions, you cannot simply multiply their derivatives. To see why, consider [latex]f(x) = x^2[\/latex]. We know [latex]f'(x) = 2x[\/latex], but if we incorrectly applied \"derivative of product = product of derivatives,\" we'd get:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{d}{dx}(x) \\cdot \\dfrac{d}{dx}(x) = 1 \\cdot 1 = 1[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This is clearly wrong. The correct rule accounts for both functions changing simultaneously.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">the product rule<\/h3>\r\n<p id=\"fs-id1169739298182\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions. Then<\/p>\r\n\r\n<div id=\"fs-id1169739326645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(f(x)g(x))=\\frac{d}{dx}(f(x))\\cdot g(x)+\\frac{d}{dx}(g(x))\\cdot f(x)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739187834\">That is,<\/p>\r\n\r\n<div id=\"fs-id1169739187837\" class=\"equation unnumbered\" style=\"text-align: center;\">if [latex]j(x)=f(x)g(x)[\/latex] then [latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739269717\">This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Remember the Pattern<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Some students find it helpful to think: \"first derivative, second stays\" + \"first stays, second derivative\"<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The order doesn't matter due to commutativity\u2014you can write the terms in either order:<\/p>\r\n<p style=\"text-align: center;\">[latex]f'(x)g(x) + g'(x)f(x) = g'(x)f(x) + f'(x)g(x)[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169736659567\">For [latex]j(x)=f(x)g(x)[\/latex], use the product rule to find [latex]j^{\\prime}(2)[\/latex] if [latex]f(2)=3, \\, f^{\\prime}(2)=-4, \\, g(2)=1[\/latex], and [latex]g^{\\prime}(2)=6[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169736658802\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736658802\"]\r\n<p id=\"fs-id1169736658802\">Since [latex]j(x)=f(x)g(x), \\, j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)[\/latex], and hence<\/p>\r\n\r\n<div id=\"fs-id1169736656614\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(2)=f^{\\prime}(2)g(2)+g^{\\prime}(2)f(2)=(-4)(1)+(6)(3)=14[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169739273822\">For [latex]j(x)=(x^2+2)(3x^3-5x)[\/latex], find [latex]j^{\\prime}(x)[\/latex] by applying the product rule. Check the result by first finding the product and then differentiating.<\/p>\r\n[reveal-answer q=\"fs-id1169739301174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739301174\"]\r\n<p id=\"fs-id1169739301174\">If we set [latex]f(x)=x^2+2[\/latex] and [latex]g(x)=3x^3-5x[\/latex], then [latex]f^{\\prime}(x)=2x[\/latex] and [latex]g^{\\prime}(x)=9x^2-5[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1169739343689\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)=(2x)(3x^3-5x)+(9x^2-5)(x^2+2)[\/latex].<\/div>\r\n<p id=\"fs-id1169736612557\">Simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1169736612560\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=15x^4+3x^2-10[\/latex].<\/div>\r\n<p id=\"fs-id1169739274892\">To check, we see that [latex]j(x)=3x^5+x^3-10x[\/latex] and, consequently, [latex]j^{\\prime}(x)=15x^4+3x^2-10[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]205283[\/ohm_question]\r\n\r\n<\/section>\r\n<h3>The Quotient Rule<\/h3>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Just as with products, you cannot simply divide derivatives when differentiating a quotient. Consider [latex]f(x) = x^2[\/latex]. We know [latex]f'(x) = 2x[\/latex], but if we incorrectly applied \"derivative of quotient = quotient of derivatives,\" we'd get:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{\\frac{d}{dx}(x^3)}{\\frac{d}{dx}(x)} = \\dfrac{3x^2}{1} = 3x^2[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">This is wrong. The quotient rule accounts for the interaction between numerator and denominator.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">the quotient rule<\/h3>\r\n<p id=\"fs-id1169736662921\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions. Then<\/p>\r\n\r\n<div id=\"fs-id1169739336009\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left(\\dfrac{f(x)}{g(x)}\\right)=\\dfrac{\\frac{d}{dx}(f(x))\\cdot g(x)-\\dfrac{d}{dx}(g(x))\\cdot f(x)}{(g(x))^2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739190663\">That is,<\/p>\r\n\r\n<div id=\"fs-id1169739190666\" class=\"equation unnumbered\" style=\"text-align: center;\">if [latex]j(x)=\\dfrac{f(x)}{g(x)}[\/latex], then [latex]j^{\\prime}(x)=\\dfrac{f^{\\prime}(x)g(x)-g^{\\prime}(x)f(x)}{(g(x))^2}[\/latex]<\/div>\r\n<\/section>\r\n<div><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Remember: \"Low-D-High minus High-D-Low, over Low-Low\"<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Many students use this mnemonic:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Low<\/strong> = denominator function [latex]g(x)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>High<\/strong> = numerator function [latex]f(x)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>D<\/strong> = derivative<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">So: [latex]\\dfrac{\\text{(Low)(D-High)} - \\text{(High)(D-Low)}}{\\text{(Low)}^2}[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Warning:<\/strong> Order matters here\u2014switching the subtraction will give you the wrong sign.<\/p>\r\n\r\n<\/section><\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739305235\">Use the quotient rule to find the derivative of [latex]k(x)=\\dfrac{5x^2}{4x+3}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739305276\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739305276\"]\r\n<p id=\"fs-id1169739305276\">Let [latex]f(x)=5x^2[\/latex] and [latex]g(x)=4x+3[\/latex]. Thus, [latex]f^{\\prime}(x)=10x[\/latex] and [latex]g^{\\prime}(x)=4[\/latex]. Substituting into the quotient rule, we have<\/p>\r\n\r\n<div id=\"fs-id1169739299200\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=\\dfrac{f^{\\prime}(x)g(x)-g^{\\prime}(x)f(x)}{(g(x))^2}=\\dfrac{10x(4x+3)-4(5x^2)}{(4x+3)^2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739299784\">Simplifying, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739299787\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=\\dfrac{20x^2+30x}{(4x+3)^2}[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169739299858\">Find the derivative of [latex]h(x)=\\dfrac{3x+1}{4x-3}[\/latex]<\/p>\r\n[reveal-answer q=\"336671\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"336671\"]\r\n<p id=\"fs-id1169739348450\">Apply the quotient rule with [latex]f(x)=3x+1[\/latex] and [latex]g(x)=4x-3[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739348394\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739348394\"]\r\n<p id=\"fs-id1169739348394\">[latex]k^{\\prime}(x)=-\\dfrac{13}{(4x-3)^2}[\/latex].<\/p>\r\nWatch the following video to see the worked solution to this example.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ruACLHzWT3g?controls=0&amp;start=931&amp;end=998&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.3DifferentiationRules931to998_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.3 Differentiation Rules\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]158799[\/ohm_question]\r\n\r\n<\/section>\r\n<h3 style=\"text-align: left;\">Extended Power Rule<\/h3>\r\n<p id=\"fs-id1169739348504\">It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form [latex]x^k[\/latex] where [latex]k[\/latex] is a negative integer.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">extended power rule<\/h3>\r\n<p id=\"fs-id1169736617603\">If [latex]k[\/latex] is a negative integer, then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{d}{dx}(x^k)=kx^{k-1}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169736614318\">Find [latex]\\frac{d}{dx}(x^{-4})[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739300028\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739300028\"]\r\n<p id=\"fs-id1169739300028\">By applying the extended power rule with [latex]k=-4[\/latex], we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739300043\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{-4})=-4x^{-4-1}=-4x^{-5}[\/latex].<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\">[\/hidden-answer]<\/div>\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169739300111\">Use the extended power rule and the constant multiple rule to find [latex]f(x)=\\dfrac{6}{x^2}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169736656728\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736656728\"]\r\n<p id=\"fs-id1169736656728\">It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do so. However, it is far easier to differentiate this function by first rewriting it as [latex]f(x)=6x^{-2}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169736656758\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}f^{\\prime}(x) &amp; =\\frac{d}{dx}(\\frac{6}{x^2})=\\frac{d}{dx}(6x^{-2}) &amp; &amp; &amp; \\text{Rewrite} \\, \\frac{6}{x^2} \\, \\text{as} \\, 6x^{-2}. \\\\ &amp; =6\\frac{d}{dx}(x^{-2}) &amp; &amp; &amp; \\text{Apply the constant multiple rule.} \\\\ &amp; =6(-2x^{-3}) &amp; &amp; &amp; \\text{Use the extended power rule to differentiate} \\, x^{-2}. \\\\ &amp; =-12x^{-3} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]33684[\/ohm_question]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Apply specific rules to find derivatives of functions multiplied or divided by each other<\/span><\/li>\n<\/ul>\n<\/section>\n<h2 class=\"entry-title\">The Advanced Rules<\/h2>\n<h3>The Product Rule<\/h3>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When differentiating a product of two functions, you cannot simply multiply their derivatives. To see why, consider [latex]f(x) = x^2[\/latex]. We know [latex]f'(x) = 2x[\/latex], but if we incorrectly applied &#8220;derivative of product = product of derivatives,&#8221; we&#8217;d get:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{d}{dx}(x) \\cdot \\dfrac{d}{dx}(x) = 1 \\cdot 1 = 1[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This is clearly wrong. The correct rule accounts for both functions changing simultaneously.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">the product rule<\/h3>\n<p id=\"fs-id1169739298182\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions. Then<\/p>\n<div id=\"fs-id1169739326645\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(f(x)g(x))=\\frac{d}{dx}(f(x))\\cdot g(x)+\\frac{d}{dx}(g(x))\\cdot f(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739187834\">That is,<\/p>\n<div id=\"fs-id1169739187837\" class=\"equation unnumbered\" style=\"text-align: center;\">if [latex]j(x)=f(x)g(x)[\/latex] then [latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739269717\">This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Remember the Pattern<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Some students find it helpful to think: &#8220;first derivative, second stays&#8221; + &#8220;first stays, second derivative&#8221;<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The order doesn&#8217;t matter due to commutativity\u2014you can write the terms in either order:<\/p>\n<p style=\"text-align: center;\">[latex]f'(x)g(x) + g'(x)f(x) = g'(x)f(x) + f'(x)g(x)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169736659567\">For [latex]j(x)=f(x)g(x)[\/latex], use the product rule to find [latex]j^{\\prime}(2)[\/latex] if [latex]f(2)=3, \\, f^{\\prime}(2)=-4, \\, g(2)=1[\/latex], and [latex]g^{\\prime}(2)=6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169736658802\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169736658802\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736658802\">Since [latex]j(x)=f(x)g(x), \\, j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)[\/latex], and hence<\/p>\n<div id=\"fs-id1169736656614\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(2)=f^{\\prime}(2)g(2)+g^{\\prime}(2)f(2)=(-4)(1)+(6)(3)=14[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739273822\">For [latex]j(x)=(x^2+2)(3x^3-5x)[\/latex], find [latex]j^{\\prime}(x)[\/latex] by applying the product rule. Check the result by first finding the product and then differentiating.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739301174\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739301174\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739301174\">If we set [latex]f(x)=x^2+2[\/latex] and [latex]g(x)=3x^3-5x[\/latex], then [latex]f^{\\prime}(x)=2x[\/latex] and [latex]g^{\\prime}(x)=9x^2-5[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169739343689\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)=(2x)(3x^3-5x)+(9x^2-5)(x^2+2)[\/latex].<\/div>\n<p id=\"fs-id1169736612557\">Simplifying, we have<\/p>\n<div id=\"fs-id1169736612560\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=15x^4+3x^2-10[\/latex].<\/div>\n<p id=\"fs-id1169739274892\">To check, we see that [latex]j(x)=3x^5+x^3-10x[\/latex] and, consequently, [latex]j^{\\prime}(x)=15x^4+3x^2-10[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm205283\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=205283&theme=lumen&iframe_resize_id=ohm205283&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<h3>The Quotient Rule<\/h3>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Just as with products, you cannot simply divide derivatives when differentiating a quotient. Consider [latex]f(x) = x^2[\/latex]. We know [latex]f'(x) = 2x[\/latex], but if we incorrectly applied &#8220;derivative of quotient = quotient of derivatives,&#8221; we&#8217;d get:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{\\frac{d}{dx}(x^3)}{\\frac{d}{dx}(x)} = \\dfrac{3x^2}{1} = 3x^2[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">This is wrong. The quotient rule accounts for the interaction between numerator and denominator.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">the quotient rule<\/h3>\n<p id=\"fs-id1169736662921\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be differentiable functions. Then<\/p>\n<div id=\"fs-id1169739336009\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left(\\dfrac{f(x)}{g(x)}\\right)=\\dfrac{\\frac{d}{dx}(f(x))\\cdot g(x)-\\dfrac{d}{dx}(g(x))\\cdot f(x)}{(g(x))^2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739190663\">That is,<\/p>\n<div id=\"fs-id1169739190666\" class=\"equation unnumbered\" style=\"text-align: center;\">if [latex]j(x)=\\dfrac{f(x)}{g(x)}[\/latex], then [latex]j^{\\prime}(x)=\\dfrac{f^{\\prime}(x)g(x)-g^{\\prime}(x)f(x)}{(g(x))^2}[\/latex]<\/div>\n<\/section>\n<div>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Remember: &#8220;Low-D-High minus High-D-Low, over Low-Low&#8221;<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Many students use this mnemonic:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Low<\/strong> = denominator function [latex]g(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>High<\/strong> = numerator function [latex]f(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>D<\/strong> = derivative<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\">So: [latex]\\dfrac{\\text{(Low)(D-High)} - \\text{(High)(D-Low)}}{\\text{(Low)}^2}[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Warning:<\/strong> Order matters here\u2014switching the subtraction will give you the wrong sign.<\/p>\n<\/section>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739305235\">Use the quotient rule to find the derivative of [latex]k(x)=\\dfrac{5x^2}{4x+3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739305276\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739305276\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739305276\">Let [latex]f(x)=5x^2[\/latex] and [latex]g(x)=4x+3[\/latex]. Thus, [latex]f^{\\prime}(x)=10x[\/latex] and [latex]g^{\\prime}(x)=4[\/latex]. Substituting into the quotient rule, we have<\/p>\n<div id=\"fs-id1169739299200\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=\\dfrac{f^{\\prime}(x)g(x)-g^{\\prime}(x)f(x)}{(g(x))^2}=\\dfrac{10x(4x+3)-4(5x^2)}{(4x+3)^2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739299784\">Simplifying, we obtain<\/p>\n<div id=\"fs-id1169739299787\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]k^{\\prime}(x)=\\dfrac{20x^2+30x}{(4x+3)^2}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739299858\">Find the derivative of [latex]h(x)=\\dfrac{3x+1}{4x-3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q336671\">Hint<\/button><\/p>\n<div id=\"q336671\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739348450\">Apply the quotient rule with [latex]f(x)=3x+1[\/latex] and [latex]g(x)=4x-3[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739348394\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739348394\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739348394\">[latex]k^{\\prime}(x)=-\\dfrac{13}{(4x-3)^2}[\/latex].<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ruACLHzWT3g?controls=0&amp;start=931&amp;end=998&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.3DifferentiationRules931to998_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.3 Differentiation Rules&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm158799\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=158799&theme=lumen&iframe_resize_id=ohm158799&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<h3 style=\"text-align: left;\">Extended Power Rule<\/h3>\n<p id=\"fs-id1169739348504\">It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form [latex]x^k[\/latex] where [latex]k[\/latex] is a negative integer.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">extended power rule<\/h3>\n<p id=\"fs-id1169736617603\">If [latex]k[\/latex] is a negative integer, then<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{d}{dx}(x^k)=kx^{k-1}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169736614318\">Find [latex]\\frac{d}{dx}(x^{-4})[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739300028\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739300028\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739300028\">By applying the extended power rule with [latex]k=-4[\/latex], we obtain<\/p>\n<div id=\"fs-id1169739300043\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{-4})=-4x^{-4-1}=-4x^{-5}[\/latex].<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739300111\">Use the extended power rule and the constant multiple rule to find [latex]f(x)=\\dfrac{6}{x^2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169736656728\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169736656728\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736656728\">It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do so. However, it is far easier to differentiate this function by first rewriting it as [latex]f(x)=6x^{-2}[\/latex].<\/p>\n<div id=\"fs-id1169736656758\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}f^{\\prime}(x) & =\\frac{d}{dx}(\\frac{6}{x^2})=\\frac{d}{dx}(6x^{-2}) & & & \\text{Rewrite} \\, \\frac{6}{x^2} \\, \\text{as} \\, 6x^{-2}. \\\\ & =6\\frac{d}{dx}(x^{-2}) & & & \\text{Apply the constant multiple rule.} \\\\ & =6(-2x^{-3}) & & & \\text{Use the extended power rule to differentiate} \\, x^{-2}. \\\\ & =-12x^{-3} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm33684\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33684&theme=lumen&iframe_resize_id=ohm33684&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-549","chapter","type-chapter","status-publish","hentry"],"part":424,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/549","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/549\/revisions"}],"predecessor-version":[{"id":2620,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/549\/revisions\/2620"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/424"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/549\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=549"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=549"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=549"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=549"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}