{"id":2871,"date":"2024-06-10T18:41:44","date_gmt":"2024-06-10T18:41:44","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2871"},"modified":"2024-08-05T12:40:53","modified_gmt":"2024-08-05T12:40:53","slug":"differentiation-rules-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/differentiation-rules-learn-it-4\/","title":{"raw":"Differentiation Rules: Learn It 4","rendered":"Differentiation Rules: Learn It 4"},"content":{"raw":"<h2 class=\"entry-title\">The Advanced Rules Cont.<\/h2>\r\n<h3>The Quotient Rule<\/h3>\r\n<p>Having developed and practiced the product rule, we now consider differentiating quotients of functions. <br \/>\r\n<br \/>\r\n<\/p>\r\n<p>As we see in the following theorem, the derivative of a quotient is not simply the quotient of the derivatives. Instead, it involves the derivative of the function in the numerator multiplied by the function in the denominator, minus the derivative of the function in the denominator multiplied by the function in the numerator, all divided by the square of the function in the denominator.<\/p>\r\n<section class=\"textbox example\">\r\n<p>To better understand why we cannot just take the quotient of the derivatives, consider that:<\/p>\r\n<div id=\"fs-id1169739269470\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^2)=2x[\/latex],<\/div>\r\n<p>which is not the same as,<\/p>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{\\frac{d}{dx}(x^3)}{\\frac{d}{dx}(x)} =\\dfrac{3x^2}{1}=3x^2[\/latex]<\/div>\r\n<\/section>\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<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<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169739190663\">That is,<\/p>\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<p id=\"fs-id1169739242563\">The proof of the <strong>quotient rule<\/strong> is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.<\/p>\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<p>[reveal-answer q=\"fs-id1169739305276\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739305276\"]<\/p>\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<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<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169739299784\">Simplifying, we obtain<\/p>\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<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>We explored the flexibility of the product rule given to the commutative property under addition and multiplication. It is worth mentioning that for the quotient rule, the order of the terms in the numerator\u00a0<strong>will<\/strong> matter, as the commutative property does not hold under subtraction. We can see this from the example above:<\/p>\r\n<p style=\"text-align: center;\">[latex]{10x(4x+3)-4(5x^2)=20x^2+30x}[\/latex]<\/p>\r\n<p>however,<\/p>\r\n<p style=\"text-align: center;\">[latex]{4(5x^2)-10x(4x+3)=-20x^2-30x}[\/latex].<\/p>\r\n<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<p>[reveal-answer q=\"336671\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"336671\"]<\/p>\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<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739348394\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739348394\"]<\/p>\r\n<p id=\"fs-id1169739348394\">[latex]k^{\\prime}(x)=-\\dfrac{13}{(4x-3)^2}[\/latex].<\/p>\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=931&amp;end=998&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\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>\r\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 \"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 tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]158799[\/ohm_question]<\/p>\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<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<\/section>\r\n<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n<hr \/>\r\n<p>If [latex]k[\/latex] is a negative integer, we may set [latex]n=\u2212k[\/latex], so that [latex]n[\/latex] is a positive integer with [latex]k=\u2212n[\/latex]. Since for each positive integer [latex]n, \\, x^{\u2212n}=\\frac{1}{x^n}[\/latex], we may now apply the quotient rule by setting [latex]f(x)=1[\/latex] and [latex]g(x)=x^n[\/latex]. In this case, [latex]f^{\\prime}(x)=0[\/latex] and [latex]g^{\\prime}(x)=nx^{n-1}[\/latex]. Thus,<\/p>\r\n<div id=\"fs-id1169739252903\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{\u2212n})=\\dfrac{0(x^n)-1(nx^{n-1})}{(x^n)^2}[\/latex].<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169739190073\">Simplifying, we see that<\/p>\r\n<div id=\"fs-id1169739190076\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{\u2212n})=\\dfrac{\u2212nx^{n-1}}{x^{2n}}=\u2212nx^{(n-1)-2n}=\u2212nx^{\u2212n-1}[\/latex].<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169736614244\">Finally, observe that since [latex]k=\u2212n[\/latex], by substituting we have<\/p>\r\n<div id=\"fs-id1169736614261\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^k)=kx^{k-1}[\/latex]<\/div>\r\n<p>[latex]_\\blacksquare[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169736614318\">Find [latex]\\frac{d}{dx}(x^{-4})[\/latex]<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739300028\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739300028\"]<\/p>\r\n<p id=\"fs-id1169739300028\">By applying the extended power rule with [latex]k=-4[\/latex], we obtain<\/p>\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>\r\n<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<p>[reveal-answer q=\"fs-id1169736656728\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169736656728\"]<\/p>\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<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\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]33684[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<h2 class=\"entry-title\">The Advanced Rules Cont.<\/h2>\n<h3>The Quotient Rule<\/h3>\n<p>Having developed and practiced the product rule, we now consider differentiating quotients of functions. <\/p>\n<p>As we see in the following theorem, the derivative of a quotient is not simply the quotient of the derivatives. Instead, it involves the derivative of the function in the numerator multiplied by the function in the denominator, minus the derivative of the function in the denominator multiplied by the function in the numerator, all divided by the square of the function in the denominator.<\/p>\n<section class=\"textbox example\">\n<p>To better understand why we cannot just take the quotient of the derivatives, consider that:<\/p>\n<div id=\"fs-id1169739269470\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^2)=2x[\/latex],<\/div>\n<p>which is not the same as,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{\\frac{d}{dx}(x^3)}{\\frac{d}{dx}(x)} =\\dfrac{3x^2}{1}=3x^2[\/latex]<\/div>\n<\/section>\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<p id=\"fs-id1169739242563\">The proof of the <strong>quotient rule<\/strong> is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.<\/p>\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<p><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<p>We explored the flexibility of the product rule given to the commutative property under addition and multiplication. It is worth mentioning that for the quotient rule, the order of the terms in the numerator\u00a0<strong>will<\/strong> matter, as the commutative property does not hold under subtraction. We can see this from the example above:<\/p>\n<p style=\"text-align: center;\">[latex]{10x(4x+3)-4(5x^2)=20x^2+30x}[\/latex]<\/p>\n<p>however,<\/p>\n<p style=\"text-align: center;\">[latex]{4(5x^2)-10x(4x+3)=-20x^2-30x}[\/latex].<\/p>\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<p><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<p><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\">\n<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><br \/>\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 connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p>If [latex]k[\/latex] is a negative integer, we may set [latex]n=\u2212k[\/latex], so that [latex]n[\/latex] is a positive integer with [latex]k=\u2212n[\/latex]. Since for each positive integer [latex]n, \\, x^{\u2212n}=\\frac{1}{x^n}[\/latex], we may now apply the quotient rule by setting [latex]f(x)=1[\/latex] and [latex]g(x)=x^n[\/latex]. In this case, [latex]f^{\\prime}(x)=0[\/latex] and [latex]g^{\\prime}(x)=nx^{n-1}[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169739252903\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{\u2212n})=\\dfrac{0(x^n)-1(nx^{n-1})}{(x^n)^2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739190073\">Simplifying, we see that<\/p>\n<div id=\"fs-id1169739190076\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{\u2212n})=\\dfrac{\u2212nx^{n-1}}{x^{2n}}=\u2212nx^{(n-1)-2n}=\u2212nx^{\u2212n-1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736614244\">Finally, observe that since [latex]k=\u2212n[\/latex], by substituting we have<\/p>\n<div id=\"fs-id1169736614261\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^k)=kx^{k-1}[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<br \/>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169736614318\">Find [latex]\\frac{d}{dx}(x^{-4})[\/latex]<\/p>\n<p><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<p><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\">\n<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><br \/>\n<\/section>\n","protected":false},"author":15,"menu_order":19,"template":"","meta":{"_candela_citation":"[]","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":[],"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\/2871"}],"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\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2871\/revisions"}],"predecessor-version":[{"id":4501,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2871\/revisions\/4501"}],"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\/2871\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=2871"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=2871"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=2871"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=2871"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}