{"id":241,"date":"2023-09-20T22:48:31","date_gmt":"2023-09-20T22:48:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/combining-differentiation-rules\/"},"modified":"2024-08-05T12:41:21","modified_gmt":"2024-08-05T12:41:21","slug":"differentiation-rules-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/differentiation-rules-learn-it-5\/","title":{"raw":"Differentiation Rules: Learn It 5","rendered":"Differentiation Rules: Learn It 5"},"content":{"raw":"<h2>Combining Differentiation Rules<\/h2>\r\n<p>As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules.\u00a0<\/p>\r\n<section class=\"textbox proTip\">\r\n<p>A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function.<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739347072\">For [latex]k(x)=3h(x)+x^2g(x)[\/latex], find [latex]k^{\\prime}(x)[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739347144\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739347144\"]<\/p>\r\n<p id=\"fs-id1169739347144\">Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.<\/p>\r\n<div id=\"fs-id1169739347147\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}k^{\\prime}(x) &amp; =\\frac{d}{dx}(3h(x)+x^2g(x))=\\frac{d}{dx}(3h(x))+\\frac{d}{dx}(x^2g(x)) &amp; &amp; &amp; \\text{Apply the sum rule.} \\\\ &amp; =3\\frac{d}{dx}(h(x))+(\\frac{d}{dx}(x^2)g(x)+\\frac{d}{dx}(g(x))x^2) &amp; &amp; &amp; \\begin{array}{l}\\text{Apply the constant multiple rule to} \\\\ \\text{differentiate} \\, 3h(x) \\, \\text{and the product} \\\\ \\text{rule to differentiate} \\, x^2g(x). \\end{array} \\\\ &amp; =3h^{\\prime}(x)+2xg(x)+g^{\\prime}(x)x^2 &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169736658401\">For [latex]h(x)=\\large \\frac{2x^3k(x)}{3x+2}[\/latex], find [latex]h^{\\prime}(x)[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169736658474\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169736658474\"]<\/p>\r\n<p id=\"fs-id1169736658474\">This procedure is typical for finding the derivative of a rational function.<\/p>\r\n<div id=\"fs-id1169736658477\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}h^{\\prime}(x) &amp; =\\large \\frac{\\frac{d}{dx}(2x^3k(x))\\cdot (3x+2)-\\frac{d}{dx}(3x+2)\\cdot (2x^3k(x))}{(3x+2)^2} &amp; &amp; &amp; \\text{Apply the quotient rule.} \\\\ &amp; =\\large \\frac{(6x^2k(x)+k^{\\prime}(x)\\cdot 2x^3)(3x+2)-3(2x^3k(x))}{(3x+2)^2} &amp; &amp; &amp; \\begin{array}{l}\\text{Apply the product rule to find} \\\\ \\frac{d}{dx}(2x^3k(x)). \\, \\text{Use} \\, \\frac{d}{dx}(3x+2)=3. \\end{array} \\\\ &amp; =\\large \\frac{-6x^3k(x)+18x^3k(x)+12x^2k(x)+6x^4k^{\\prime}(x)+4x^3k^{\\prime}(x)}{(3x+2)^2} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169736589245\">Determine the values of [latex]x[\/latex] for which [latex]f(x)=x^3-7x^2+8x+1[\/latex] has a horizontal tangent line.<\/p>\r\n<p>[reveal-answer q=\"fs-id1169736589298\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169736589298\"]<\/p>\r\n<p id=\"fs-id1169736589298\">To find the values of [latex]x[\/latex] for which [latex]f(x)[\/latex] has a horizontal tangent line, we must solve [latex]f^{\\prime}(x)=0[\/latex]. Since<\/p>\r\n<div id=\"fs-id1169736589343\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=3x^2-14x+8=(3x-2)(x-4)[\/latex],<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169739111144\">we must solve [latex](3x-2)(x-4)=0[\/latex]. Thus we see that the function has horizontal tangent lines at [latex]x=\\frac{2}{3}[\/latex] and [latex]x=4[\/latex] as shown in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"379\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205334\/CNX_Calc_Figure_03_03_002.jpg\" alt=\"The graph shows f(x) = x3 \u2013 7x2 + 8x + 1, and the tangent lines are shown as x = 2\/3 and x = 4.\" width=\"379\" height=\"310\" \/> Figure 2. This function has horizontal tangent lines at [latex]x = 2\/3[\/latex] and [latex]x = 4[\/latex].[\/caption]\r\n\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ruACLHzWT3g?controls=0&amp;start=1476&amp;end=1576&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p><br \/>\r\nFor 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>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.3DifferentiationRules1476to1576_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.3 Differentiation Rules\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739281986\">The position of an object on a coordinate axis at time [latex]t[\/latex] is given by [latex]s(t)=\\dfrac{t}{t^2+1}[\/latex].<\/p>\r\n<p>What is the initial velocity of the object?<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739282028\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739282028\"]<\/p>\r\n<p id=\"fs-id1169739282028\">Since the initial velocity is [latex]v(0)=s^{\\prime}(0)[\/latex], begin by finding [latex]s^{\\prime}(t)[\/latex] by applying the quotient rule:<\/p>\r\n<div id=\"fs-id1169739282080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s^{\\prime}(t)=\\dfrac{1(t^2+1)-2t(t)}{(t^2+1)^2}=\\dfrac{1-t^2}{(t^2+1)^2}[\/latex].<\/div>\r\n<p id=\"fs-id1169739301434\">After evaluating, we see that [latex]v(0)=1[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]33700[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<h2>Combining Differentiation Rules<\/h2>\n<p>As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules.\u00a0<\/p>\n<section class=\"textbox proTip\">\n<p>A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739347072\">For [latex]k(x)=3h(x)+x^2g(x)[\/latex], find [latex]k^{\\prime}(x)[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739347144\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739347144\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739347144\">Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.<\/p>\n<div id=\"fs-id1169739347147\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}k^{\\prime}(x) & =\\frac{d}{dx}(3h(x)+x^2g(x))=\\frac{d}{dx}(3h(x))+\\frac{d}{dx}(x^2g(x)) & & & \\text{Apply the sum rule.} \\\\ & =3\\frac{d}{dx}(h(x))+(\\frac{d}{dx}(x^2)g(x)+\\frac{d}{dx}(g(x))x^2) & & & \\begin{array}{l}\\text{Apply the constant multiple rule to} \\\\ \\text{differentiate} \\, 3h(x) \\, \\text{and the product} \\\\ \\text{rule to differentiate} \\, x^2g(x). \\end{array} \\\\ & =3h^{\\prime}(x)+2xg(x)+g^{\\prime}(x)x^2 & & & \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169736658401\">For [latex]h(x)=\\large \\frac{2x^3k(x)}{3x+2}[\/latex], find [latex]h^{\\prime}(x)[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169736658474\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169736658474\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736658474\">This procedure is typical for finding the derivative of a rational function.<\/p>\n<div id=\"fs-id1169736658477\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}h^{\\prime}(x) & =\\large \\frac{\\frac{d}{dx}(2x^3k(x))\\cdot (3x+2)-\\frac{d}{dx}(3x+2)\\cdot (2x^3k(x))}{(3x+2)^2} & & & \\text{Apply the quotient rule.} \\\\ & =\\large \\frac{(6x^2k(x)+k^{\\prime}(x)\\cdot 2x^3)(3x+2)-3(2x^3k(x))}{(3x+2)^2} & & & \\begin{array}{l}\\text{Apply the product rule to find} \\\\ \\frac{d}{dx}(2x^3k(x)). \\, \\text{Use} \\, \\frac{d}{dx}(3x+2)=3. \\end{array} \\\\ & =\\large \\frac{-6x^3k(x)+18x^3k(x)+12x^2k(x)+6x^4k^{\\prime}(x)+4x^3k^{\\prime}(x)}{(3x+2)^2} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169736589245\">Determine the values of [latex]x[\/latex] for which [latex]f(x)=x^3-7x^2+8x+1[\/latex] has a horizontal tangent line.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169736589298\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169736589298\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736589298\">To find the values of [latex]x[\/latex] for which [latex]f(x)[\/latex] has a horizontal tangent line, we must solve [latex]f^{\\prime}(x)=0[\/latex]. Since<\/p>\n<div id=\"fs-id1169736589343\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=3x^2-14x+8=(3x-2)(x-4)[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739111144\">we must solve [latex](3x-2)(x-4)=0[\/latex]. Thus we see that the function has horizontal tangent lines at [latex]x=\\frac{2}{3}[\/latex] and [latex]x=4[\/latex] as shown in the following graph.<\/p>\n<figure style=\"width: 379px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205334\/CNX_Calc_Figure_03_03_002.jpg\" alt=\"The graph shows f(x) = x3 \u2013 7x2 + 8x + 1, and the tangent lines are shown as x = 2\/3 and x = 4.\" width=\"379\" height=\"310\" \/><figcaption class=\"wp-caption-text\">Figure 2. This function has horizontal tangent lines at [latex]x = 2\/3[\/latex] and [latex]x = 4[\/latex].<\/figcaption><\/figure>\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=1476&amp;end=1576&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>\nFor 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\/3.3DifferentiationRules1476to1576_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 example\">\n<p id=\"fs-id1169739281986\">The position of an object on a coordinate axis at time [latex]t[\/latex] is given by [latex]s(t)=\\dfrac{t}{t^2+1}[\/latex].<\/p>\n<p>What is the initial velocity of the object?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739282028\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739282028\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739282028\">Since the initial velocity is [latex]v(0)=s^{\\prime}(0)[\/latex], begin by finding [latex]s^{\\prime}(t)[\/latex] by applying the quotient rule:<\/p>\n<div id=\"fs-id1169739282080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]s^{\\prime}(t)=\\dfrac{1(t^2+1)-2t(t)}{(t^2+1)^2}=\\dfrac{1-t^2}{(t^2+1)^2}[\/latex].<\/div>\n<p id=\"fs-id1169739301434\">After evaluating, we see that [latex]v(0)=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm33700\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33700&theme=lumen&iframe_resize_id=ohm33700&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":20,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.3 Differentiation Rules\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":503,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"3.3 Differentiation Rules","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/241"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/241\/revisions"}],"predecessor-version":[{"id":4503,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/241\/revisions\/4503"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/503"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/241\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=241"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=241"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=241"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=241"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}