{"id":338,"date":"2021-02-04T01:14:51","date_gmt":"2021-02-04T01:14:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=338"},"modified":"2022-03-16T05:29:44","modified_gmt":"2022-03-16T05:29:44","slug":"the-advanced-rules","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/the-advanced-rules\/","title":{"raw":"The Advanced Rules","rendered":"The Advanced Rules"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the product rule for finding the derivative of a product of functions<\/li>\r\n \t<li>Use the quotient rule for finding the derivative of a quotient of functions<\/li>\r\n \t<li>Extend the power rule to functions with negative exponents<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Product Rule<\/h2>\r\n<p id=\"fs-id1169739194716\">Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the <strong>product rule<\/strong> does not follow this pattern. To see why we cannot use this pattern, consider the function [latex]f(x)=x^2[\/latex], whose derivative is [latex]f^{\\prime}(x)=2x[\/latex] and not [latex]\\frac{d}{dx}(x)\\cdot \\frac{d}{dx}(x)=1\\cdot 1=1[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169739298175\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Product Rule<\/h3>\r\n\r\n<hr \/>\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<div><\/div>\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<div><\/div>\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<\/div>\r\n<div id=\"fs-id1169736654322\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\n<p id=\"fs-id1169736654327\">We begin by assuming that [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are differentiable functions. At a key point in this proof we need to use the fact that, since [latex]g(x)[\/latex] is differentiable, it is also continuous. In particular, we use the fact that since [latex]g(x)[\/latex] is continuous, [latex]\\underset{h\\to 0}{\\lim}g(x+h)=g(x)[\/latex].<\/p>\r\n<p id=\"fs-id1169736656513\">By applying the limit definition of the derivative to [latex]j(x)=f(x)g(x)[\/latex], we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739327881\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739301642\">By adding and subtracting [latex]f(x)g(x+h)[\/latex] in the numerator, we have<\/p>\r\n\r\n<div id=\"fs-id1169739301672\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739304787\">After breaking apart this quotient and applying the sum law for limits, the derivative becomes<\/p>\r\n\r\n<div id=\"fs-id1169739304791\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\left(\\frac{f(x+h)g(x+h)-f(x)g(x+h)}{h}\\right)+\\underset{h\\to 0}{\\lim}\\left(\\frac{f(x)g(x+h)-f(x)g(x)}{h}\\right)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736662807\">Rearranging, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169736662810\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\left(\\frac{f(x+h)-f(x)}{h}\\cdot g(x+h)\\right)+\\underset{h\\to 0}{\\lim}\\left(\\frac{g(x+h)-g(x)}{h}\\cdot f(x)\\right)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738916876\">By using the continuity of [latex]g(x)[\/latex], the definition of the derivatives of [latex]f(x)[\/latex] and [latex]g(x)[\/latex], and applying the limit laws, we arrive at the product rule,<\/p>\r\n\r\n<div id=\"fs-id1169736655920\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)[\/latex]<\/div>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\nAs you begin using the product rule, it may be useful to remember that addition and multiplication are commutative for all real numbers.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: commutative property of addition and multiplication for real numbers<\/h3>\r\nThe following properties hold for real numbers <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>.\r\n<table summary=\"A table with six rows and three columns. The first entry of the first row is blank while the remaining columns read: Addition and Multiplication. The first entry of the second row reads: Commutative Property. The second column entry reads a plus b equals b plus a. The third column entry reads a times b equals b times a. The first entry of the third row reads Associative Property. The second column entry reads: a plus the quantity b plus c in parenthesis equals the quantity a plus b in parenthesis plus c. The third column entry reads: a times the quantity b times c in parenthesis equals the quantity a times b in parenthesis times c. The first entry of the fourth row reads: Distributive Property. The second and third column are combined on this row and read: a times the quantity b plus c in parenthesis equals a times b plus a times c. The first entry in the fifth row reads: Identity Property. The second column entry reads: There exists a unique real number called the additive identity, 0, such that for any real number a, a + 0 = a. The third column entry reads: There exists a unique real number called the multiplicative inverse, 1, such that for any real number a, a times 1 equals a. The first entry in the sixth row reads: Inverse Property. The second column entry reads: Every real number a has an additive inverse, or opposite, denoted negative a such that, a plus negative a equals zero. The third column entry reads: Every nonzero real\">\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>Addition<\/th>\r\n<th>Multiplication<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td><strong>Commutative Property<\/strong><\/td>\r\n<td>[latex]a+b=b+a[\/latex]<\/td>\r\n<td>[latex]a\\cdot b=b\\cdot a[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nThis is particularly useful for our product rule because our formula consists solely of these two operations. Due to the commutative property of addition:\r\n<p style=\"text-align: center;\">[latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)=g^{\\prime}(x)f(x)+f^{\\prime}(x)g(x)[\/latex]<\/p>\r\nAdditionally, the order in which you multiply each of these terms doesn't matter, due to the commutative property of multiplication.\r\n<div id=\"fs-id1169736659557\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Product Rule to Functions at a Point<\/h3>\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<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739273812\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Product Rule to Binomials<\/h3>\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&nbsp;\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&nbsp;\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<\/div>\r\n<div id=\"fs-id1169736654821\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169736654828\">Use the product rule to obtain the derivative of [latex]j(x)=2x^5(4x^2+x)[\/latex].<\/p>\r\n[reveal-answer q=\"034256\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"034256\"]\r\n<p id=\"fs-id1169736609959\">Set [latex]f(x)=2x^5[\/latex] and [latex]g(x)=4x^2+x[\/latex] and use the preceding example as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169736654876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736654876\"]\r\n<p id=\"fs-id1169736654876\">[latex]j^{\\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ruACLHzWT3g?controls=0&amp;start=685&amp;end=751&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]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\/3.3DifferentiationRules685to751_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.3 Differentiation Rules\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]205283[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>The Quotient Rule<\/h2>\r\n<p id=\"fs-id1169739269461\">Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that<\/p>\r\n\r\n<div id=\"fs-id1169739269470\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^2)=2x[\/latex], which is not the same as [latex]\\dfrac{\\frac{d}{dx}(x^3)}{\\frac{d}{dx}(x)} =\\dfrac{3x^2}{1}=3x^2[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div id=\"fs-id1169736662915\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">The Quotient Rule<\/h3>\r\n\r\n<hr \/>\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<div><\/div>\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&nbsp;\r\n\r\n<\/div>\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\r\n<div id=\"fs-id1169739305225\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Quotient Rule<\/h3>\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\">[latex]k^{\\prime}(x)=\\dfrac{20x^2+30x}{(4x+3)^2}[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe 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:\r\n<p style=\"text-align: center;\">[latex]{10x(4x+3)-4(5x^2)=20x^2+30x}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">however,<\/p>\r\n<p style=\"text-align: center;\">[latex]{4(5x^2)-10x(4x+3)=-20x^2-30x}[\/latex]<\/p>\r\n&nbsp;\r\n<div id=\"fs-id1169739299850\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\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\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\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\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]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>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]158799[\/ohm_question]\r\n\r\n<\/div>\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<div id=\"fs-id1169736617597\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Extended Power Rule<\/h3>\r\n\r\n<hr \/>\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&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1169736617655\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\r\nIf [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,\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&nbsp;\r\n<p id=\"fs-id1169739190073\">Simplifying, we see that<\/p>\r\n\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&nbsp;\r\n<p id=\"fs-id1169736614244\">Finally, observe that since [latex]k=\u2212n[\/latex], by substituting we have<\/p>\r\n\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[latex]_\\blacksquare[\/latex]\r\n<div><\/div>\r\n<div id=\"fs-id1169736614308\" class=\"textbook exercises\">\r\n<h3>Example: Using the Extended Power Rule<\/h3>\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&nbsp;\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739300101\" class=\"textbook exercises\">\r\n<h3>Example: Using the Extended Power Rule and the Constant Multiple Rule<\/h3>\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&nbsp;\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169736657086\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739251962\">Find the derivative of [latex]g(x)=\\dfrac{1}{x^7}[\/latex] using the extended power rule.<\/p>\r\n[reveal-answer q=\"873564\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"873564\"]\r\n<p id=\"fs-id1169739252030\">Rewrite [latex]g(x)=\\frac{1}{x^7}=x^{-7}[\/latex]. Use the extended power rule with [latex]k=-7[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739251993\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739251993\"]\r\n<p id=\"fs-id1169739251993\">[latex]g^{\\prime}(x)=-7x^{-8}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/ruACLHzWT3g?controls=0&amp;start=1125&amp;end=1148&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]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\/3.3DifferentiationRules1125to1148_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.3 Differentiation Rules\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]33684[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the product rule for finding the derivative of a product of functions<\/li>\n<li>Use the quotient rule for finding the derivative of a quotient of functions<\/li>\n<li>Extend the power rule to functions with negative exponents<\/li>\n<\/ul>\n<\/div>\n<h2>The Product Rule<\/h2>\n<p id=\"fs-id1169739194716\">Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the <strong>product rule<\/strong> does not follow this pattern. To see why we cannot use this pattern, consider the function [latex]f(x)=x^2[\/latex], whose derivative is [latex]f^{\\prime}(x)=2x[\/latex] and not [latex]\\frac{d}{dx}(x)\\cdot \\frac{d}{dx}(x)=1\\cdot 1=1[\/latex].<\/p>\n<div id=\"fs-id1169739298175\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Product Rule<\/h3>\n<hr \/>\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<div><\/div>\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<div><\/div>\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<\/div>\n<div id=\"fs-id1169736654322\" class=\"bc-section section\">\n<h3>Proof<\/h3>\n<p id=\"fs-id1169736654327\">We begin by assuming that [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are differentiable functions. At a key point in this proof we need to use the fact that, since [latex]g(x)[\/latex] is differentiable, it is also continuous. In particular, we use the fact that since [latex]g(x)[\/latex] is continuous, [latex]\\underset{h\\to 0}{\\lim}g(x+h)=g(x)[\/latex].<\/p>\n<p id=\"fs-id1169736656513\">By applying the limit definition of the derivative to [latex]j(x)=f(x)g(x)[\/latex], we obtain<\/p>\n<div id=\"fs-id1169739327881\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739301642\">By adding and subtracting [latex]f(x)g(x+h)[\/latex] in the numerator, we have<\/p>\n<div id=\"fs-id1169739301672\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739304787\">After breaking apart this quotient and applying the sum law for limits, the derivative becomes<\/p>\n<div id=\"fs-id1169739304791\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\left(\\frac{f(x+h)g(x+h)-f(x)g(x+h)}{h}\\right)+\\underset{h\\to 0}{\\lim}\\left(\\frac{f(x)g(x+h)-f(x)g(x)}{h}\\right)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736662807\">Rearranging, we obtain<\/p>\n<div id=\"fs-id1169736662810\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\left(\\frac{f(x+h)-f(x)}{h}\\cdot g(x+h)\\right)+\\underset{h\\to 0}{\\lim}\\left(\\frac{g(x+h)-g(x)}{h}\\cdot f(x)\\right)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738916876\">By using the continuity of [latex]g(x)[\/latex], the definition of the derivatives of [latex]f(x)[\/latex] and [latex]g(x)[\/latex], and applying the limit laws, we arrive at the product rule,<\/p>\n<div id=\"fs-id1169736655920\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p>As you begin using the product rule, it may be useful to remember that addition and multiplication are commutative for all real numbers.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: commutative property of addition and multiplication for real numbers<\/h3>\n<p>The following properties hold for real numbers <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>.<\/p>\n<table summary=\"A table with six rows and three columns. The first entry of the first row is blank while the remaining columns read: Addition and Multiplication. The first entry of the second row reads: Commutative Property. The second column entry reads a plus b equals b plus a. The third column entry reads a times b equals b times a. The first entry of the third row reads Associative Property. The second column entry reads: a plus the quantity b plus c in parenthesis equals the quantity a plus b in parenthesis plus c. The third column entry reads: a times the quantity b times c in parenthesis equals the quantity a times b in parenthesis times c. The first entry of the fourth row reads: Distributive Property. The second and third column are combined on this row and read: a times the quantity b plus c in parenthesis equals a times b plus a times c. The first entry in the fifth row reads: Identity Property. The second column entry reads: There exists a unique real number called the additive identity, 0, such that for any real number a, a + 0 = a. The third column entry reads: There exists a unique real number called the multiplicative inverse, 1, such that for any real number a, a times 1 equals a. The first entry in the sixth row reads: Inverse Property. The second column entry reads: Every real number a has an additive inverse, or opposite, denoted negative a such that, a plus negative a equals zero. The third column entry reads: Every nonzero real\">\n<thead>\n<tr>\n<th><\/th>\n<th>Addition<\/th>\n<th>Multiplication<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><strong>Commutative Property<\/strong><\/td>\n<td>[latex]a+b=b+a[\/latex]<\/td>\n<td>[latex]a\\cdot b=b\\cdot a[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>This is particularly useful for our product rule because our formula consists solely of these two operations. Due to the commutative property of addition:<\/p>\n<p style=\"text-align: center;\">[latex]j^{\\prime}(x)=f^{\\prime}(x)g(x)+g^{\\prime}(x)f(x)=g^{\\prime}(x)f(x)+f^{\\prime}(x)g(x)[\/latex]<\/p>\n<p>Additionally, the order in which you multiply each of these terms doesn&#8217;t matter, due to the commutative property of multiplication.<\/p>\n<div id=\"fs-id1169736659557\" class=\"textbook exercises\">\n<h3>Example: Applying the Product Rule to Functions at a Point<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736658802\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<div id=\"fs-id1169739273812\" class=\"textbook exercises\">\n<h3>Example: Applying the Product Rule to Binomials<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739301174\">Show Solution<\/span><\/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>&nbsp;<\/p>\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>&nbsp;<\/p>\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<\/div>\n<div id=\"fs-id1169736654821\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169736654828\">Use the product rule to obtain the derivative of [latex]j(x)=2x^5(4x^2+x)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q034256\">Hint<\/span><\/p>\n<div id=\"q034256\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736609959\">Set [latex]f(x)=2x^5[\/latex] and [latex]g(x)=4x^2+x[\/latex] and use the preceding example as a guide.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736654876\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736654876\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736654876\">[latex]j^{\\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/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=685&amp;end=751&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">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\/3.3DifferentiationRules685to751_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.3 Differentiation Rules&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm205283\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=205283&theme=oea&iframe_resize_id=ohm205283&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>The Quotient Rule<\/h2>\n<p id=\"fs-id1169739269461\">Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that<\/p>\n<div id=\"fs-id1169739269470\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^2)=2x[\/latex], which is not the same as [latex]\\dfrac{\\frac{d}{dx}(x^3)}{\\frac{d}{dx}(x)} =\\dfrac{3x^2}{1}=3x^2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div id=\"fs-id1169736662915\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">The Quotient Rule<\/h3>\n<hr \/>\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<div><\/div>\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<p>&nbsp;<\/p>\n<\/div>\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<div id=\"fs-id1169739305225\" class=\"textbook exercises\">\n<h3>Example: Applying the Quotient Rule<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739305276\">Show Solution<\/span><\/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\">[latex]k^{\\prime}(x)=\\dfrac{20x^2+30x}{(4x+3)^2}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\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 style=\"text-align: center;\">however,<\/p>\n<p style=\"text-align: center;\">[latex]{4(5x^2)-10x(4x+3)=-20x^2-30x}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169739299850\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q336671\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739348394\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/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\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">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>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm158799\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=158799&theme=oea&iframe_resize_id=ohm158799&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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<div id=\"fs-id1169736617597\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Extended Power Rule<\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1169736617655\" class=\"bc-section section\">\n<h3>Proof<\/h3>\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]<\/p>\n<div><\/div>\n<div id=\"fs-id1169736614308\" class=\"textbook exercises\">\n<h3>Example: Using the Extended Power Rule<\/h3>\n<p id=\"fs-id1169736614318\">Find [latex]\\frac{d}{dx}(x^{-4})[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739300028\">Show Solution<\/span><\/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<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739300101\" class=\"textbook exercises\">\n<h3>Example: Using the Extended Power Rule and the Constant Multiple Rule<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736656728\">Show Solution<\/span><\/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<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736657086\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739251962\">Find the derivative of [latex]g(x)=\\dfrac{1}{x^7}[\/latex] using the extended power rule.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q873564\">Hint<\/span><\/p>\n<div id=\"q873564\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739252030\">Rewrite [latex]g(x)=\\frac{1}{x^7}=x^{-7}[\/latex]. Use the extended power rule with [latex]k=-7[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739251993\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739251993\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739251993\">[latex]g^{\\prime}(x)=-7x^{-8}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/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=1125&amp;end=1148&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" class=\"hidden-answer\" style=\"display: none\">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\/3.3DifferentiationRules1125to1148_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.3 Differentiation Rules&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm33684\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33684&theme=oea&iframe_resize_id=ohm33684&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-338\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.3 Differentiation Rules. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":13,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-338","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/338","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":38,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/338\/revisions"}],"predecessor-version":[{"id":4807,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/338\/revisions\/4807"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/35"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/338\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=338"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=338"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=338"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=338"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}