{"id":683,"date":"2025-06-20T17:02:18","date_gmt":"2025-06-20T17:02:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=683"},"modified":"2025-08-28T12:37:00","modified_gmt":"2025-08-28T12:37:00","slug":"advanced-integration-techniques-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/advanced-integration-techniques-background-youll-need-2\/","title":{"raw":"Advanced Integration Techniques: Background You'll Need 2","rendered":"Advanced Integration Techniques: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Apply differentiation formulas to find derivatives of logarithmic functions and inverse trigonometric functions<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivative of the Logarithmic Function<\/h2>\r\n<p id=\"fs-id1169738222225\">We can use implicit differentiation to find the derivative of the natural logarithmic function by working with its relationship to the natural exponential function.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">derivative of the natural logarithmic function<\/h3>\r\n<p id=\"fs-id1169737911417\">If [latex]x&gt;0[\/latex] and [latex]y=\\ln x[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1169738223534\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737765282\">More generally, let [latex]g(x)[\/latex] be a differentiable function. For all values of [latex]x[\/latex] for which [latex]g^{\\prime}(x)&gt;0[\/latex], the derivative of [latex]h(x)=\\ln(g(x))[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1169737919348\" class=\"equation\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{g(x)} g^{\\prime}(x)[\/latex]<\/div>\r\n<\/section>The graph of [latex]y=\\ln x[\/latex] and its derivative [latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex] are shown in Figure 3.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205536\/CNX_Calc_Figure_03_09_003.jpg\" alt=\"Graph of the function ln x along with its derivative 1\/x. The function ln x is increasing on (0, + \u221e). Its derivative is decreasing but greater than 0 on (0, + \u221e).\" width=\"487\" height=\"209\" \/> Figure 3.\u00a0The function [latex]y=\\ln x[\/latex] is increasing on [latex](0,+\\infty)[\/latex]. Its derivative [latex]y^{\\prime} =\\frac{1}{x}[\/latex] is greater than zero on [latex](0,+\\infty)[\/latex].[\/caption]<section class=\"textbox example\">\r\n<p id=\"fs-id1169738228377\">Find the derivative of [latex]f(x)=\\ln(x^3+3x-4)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169738221312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738221312\"]\r\n<p id=\"fs-id1169738221312\">Use the derivative of a natural logarithm directly.<\/p>\r\n\r\n<div id=\"fs-id1169738220266\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} f^{\\prime}(x) &amp; =\\frac{1}{x^3+3x-4} \\cdot (3x^2+3) &amp; &amp; &amp; \\text{Use} \\, g(x)=x^3+3x-4 \\, \\text{in} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ &amp; =\\frac{3x^2+3}{x^3+3x-4} &amp; &amp; &amp; \\text{Rewrite.} \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169738106144\">Find the derivative of [latex]f(x)=\\ln\\left(\\dfrac{x^2 \\sin x}{2x+1}\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169738219674\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738219674\"]\r\n<p id=\"fs-id1169738219674\">At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.<\/p>\r\n\r\n<div id=\"fs-id1169738219679\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} f(x) &amp; = \\ln(\\frac{x^2 \\sin x}{2x+1})=2\\ln x+\\ln(\\sin x)-\\ln(2x+1) &amp; &amp; &amp; \\text{Apply properties of logarithms.} \\\\ f^{\\prime}(x) &amp; = \\frac{2}{x} + \\frac{\\cos x}{\\sin x} -\\frac{2}{2x+1} &amp; &amp; &amp; \\text{Apply sum rule and} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ &amp; = \\frac{2}{x} + \\cot x - \\frac{2}{2x+1} &amp; &amp; &amp; \\text{Simplify using the quotient identity for cotangent.} \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]<\/section>\r\n<p id=\"fs-id1169738076342\">Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of [latex]y=\\log_b x[\/latex] and [latex]y=b^x[\/latex] for [latex]b&gt;0, \\, b\\ne 1[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">derivatives of general exponential and logarithmic functions<\/h3>\r\n<p id=\"fs-id1169738186170\">Let [latex]b&gt;0, \\, b\\ne 1[\/latex], and let [latex]g(x)[\/latex] be a differentiable function.<\/p>\r\n\r\n<ol id=\"fs-id1169737998025\">\r\n \t<li>If [latex]y=\\log_b x[\/latex], then\r\n<div id=\"fs-id1169738105090\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x \\ln b}[\/latex]<\/div>\r\nMore generally, if [latex]h(x)=\\log_b (g(x))[\/latex], then for all values of [latex]x[\/latex] for which [latex]g(x)&gt;0[\/latex],\r\n<div id=\"fs-id1169738186308\" class=\"equation\">\r\n<p style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{g^{\\prime}(x)}{g(x) \\ln b}[\/latex]<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>If [latex]y=b^x[\/latex], then\r\n<div id=\"fs-id1169738224034\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=b^x \\ln b[\/latex]<\/div>\r\nMore generally, if [latex]h(x)=b^{g(x)}[\/latex], then\r\n<div id=\"fs-id1169738045159\" class=\"equation\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=b^{g(x)} g^{\\prime}(x) \\ln b[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169738185259\">Find the derivative of [latex]h(x)= \\dfrac{3^x}{3^x+2}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169737700313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737700313\"]\r\n<p id=\"fs-id1169737700313\">Use the quotient rule and the derivative from above.<\/p>\r\n\r\n<div id=\"fs-id1169737700320\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} h^{\\prime}(x) &amp; = \\large \\frac{3^x \\ln 3(3^x+2)-3^x \\ln 3(3^x)}{(3^x+2)^2} &amp; &amp; &amp; \\text{Apply the quotient rule.} \\\\ &amp; = \\large \\frac{2 \\cdot 3^x \\ln 3}{(3^x+2)^2} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169738219403\">Find the slope of the line tangent to the graph of [latex]y=\\log_2 (3x+1)[\/latex] at [latex]x=1[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169738045067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738045067\"]\r\n<p id=\"fs-id1169738045067\">To find the slope, we must evaluate [latex]\\dfrac{dy}{dx}[\/latex] at [latex]x=1[\/latex]. Using the derivative above, we see that<\/p>\r\n\r\n<div id=\"fs-id1169738197861\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{dy}{dx}=\\dfrac{3}{\\ln 2(3x+1)}[\/latex]<\/div>\r\n<p id=\"fs-id1169738186987\">By evaluating the derivative at [latex]x=1[\/latex], we see that the tangent line has slope<\/p>\r\n\r\n<div id=\"fs-id1169738187002\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}|_{x=1} =\\frac{3}{4 \\ln 2}=\\frac{3}{\\ln 16}[\/latex]<\/div>\r\n[\/hidden-answer]<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]288388[\/ohm_question]\r\n\r\n<\/section>\r\n<h2>Derivatives of Inverse Trigonometric Functions<\/h2>\r\n<p id=\"fs-id1169739029363\">The derivatives of inverse trigonometric functions play a crucial role in the study of integration and reveal a fascinating mathematical pattern. Unlike their trigonometric counterparts, the derivatives of inverse trigonometric functions are algebraic rather than trigonometric. This illustrates an important mathematical insight: the derivative of a function does not necessarily share the same type as the original function.<\/p>\r\n\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739188589\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\sin^{-1} x[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169739269789\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739269789\"]\r\n<p id=\"fs-id1169739269789\">Since for [latex]x[\/latex] in the interval [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}], \\, f(x)= \\sin x[\/latex] is the inverse of [latex]g(x)= \\sin^{-1} x[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex].<\/p>\r\nSince:\r\n<div id=\"fs-id1169736611678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)= \\cos x[\/latex] and [latex]f^{\\prime}(g(x))= \\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex],<\/div>\r\n<p id=\"fs-id1169739347090\">we see that:<\/p>\r\n\r\n<div id=\"fs-id1169739369239\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{d}{dx}(\\sin^{-1} x)=\\frac{1}{f^{\\prime}(g(x))}=\\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/div>\r\n<strong>Analysis<\/strong>\r\n<p id=\"fs-id1169739111057\">To see that [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex], consider the following argument. Set [latex]\\sin^{-1} x=\\theta[\/latex]. In this case, [latex]\\sin \\theta =x[\/latex] where [latex]-\\frac{\\pi}{2}\\le \\theta \\le \\frac{\\pi}{2}[\/latex]. We begin by considering the case where [latex]0&lt;\\theta &lt;\\frac{\\pi}{2}[\/latex]. Since [latex]\\theta[\/latex] is an acute angle, we may construct a right triangle having acute angle [latex]\\theta[\/latex], a hypotenuse of length 1, and the side opposite angle [latex]\\theta [\/latex] having length [latex]x[\/latex]. From the Pythagorean theorem, the side adjacent to angle [latex]\\theta[\/latex] has length [latex]\\sqrt{1-x^2}[\/latex]. This triangle is shown in Figure 2. Using the triangle, we see that [latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"426\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205436\/CNX_Calc_Figure_03_07_002.jpg\" alt=\"A right triangle with angle \u03b8, opposite side x, hypotenuse 1, and adjacent side equal to the square root of the quantity (1 \u2013 x2).\" width=\"426\" height=\"205\" \/> Figure 2. Using a right triangle having acute angle [latex]\\theta[\/latex], a hypotenuse of length 1, and the side opposite angle [latex]\\theta[\/latex] having length [latex]x[\/latex], we can see that [latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex].[\/caption]\r\n<div id=\"fs-id1169739303300\" class=\"equation unnumbered\">In the case where [latex]-\\frac{\\pi}{2}&lt;\\theta &lt;0[\/latex], we make the observation that [latex]0&lt;-\\theta&lt;\\frac{\\pi}{2}[\/latex] and hence [latex]\\cos (\\sin^{-1} x)= \\cos \\theta = \\cos (\u2212\\theta )=\\sqrt{1-x^2}[\/latex].<\/div>\r\n<p id=\"fs-id1169736611289\">Now if [latex]\\theta =\\frac{\\pi}{2}[\/latex] or [latex]\\theta =-\\frac{\\pi}{2}, \\, x=1[\/latex] or [latex]x=-1[\/latex], and since in either case [latex]\\cos \\theta =0[\/latex] and [latex]\\sqrt{1-x^2}=0[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1169739340290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex]<\/div>\r\n<p id=\"fs-id1169739190632\">Consequently, in all cases, [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169739190553\">Apply the chain rule to find the derivative of [latex]h(x)=\\sin^{-1} (g(x))[\/latex] and use this result to find the derivative of [latex]h(x)=\\sin^{-1}(2x^3)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169739189899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739189899\"]\r\n<p id=\"fs-id1169739189899\">Applying the chain rule to [latex]h(x)=\\sin^{-1} (g(x))[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1169736596017\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{\\sqrt{1-(g(x))^2}}g^{\\prime}(x)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739269950\">Now let [latex]g(x)=2x^3[\/latex], so [latex]g^{\\prime}(x)=6x^{2}[\/latex]. Substituting into the previous result, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739336080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} h^{\\prime}(x) &amp; =\\dfrac{1}{\\sqrt{1-4x^6}} \\cdot 6x^{2} \\\\ &amp; =\\dfrac{6x^{2}}{\\sqrt{1-4x^6}} \\end{array}[\/latex]<\/div>\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\/0dlA5QJZYsw?controls=0&amp;start=628&amp;end=723&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\/3.7DerivativesOfInverseFunctions628to723_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.7 Derivatives of Inverse Functions (edited)\" here (opens in new window)<\/a>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<p id=\"fs-id1169739282743\">The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">derivatives of inverse trigonometric functions<\/h3>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\frac{d}{dx}(\\sin^{-1} x)=\\large \\frac{1}{\\sqrt{1-x^2}} &amp; &amp; &amp; &amp; \\frac{d}{dx}(\\cos^{-1} x)=\\large \\frac{-1}{\\sqrt{1-x^2}} \\\\ \\frac{d}{dx}(\\tan^{-1} x)=\\large \\frac{1}{1+x^2} &amp; &amp; &amp; &amp; \\frac{d}{dx}(\\cot^{-1} x)=\\large \\frac{-1}{1+x^2} \\\\ \\frac{d}{dx}(\\sec^{-1} x)=\\large \\frac{1}{|x|\\sqrt{x^2-1}} &amp; &amp; &amp; &amp; \\frac{d}{dx}(\\csc^{-1} x)=\\large \\frac{-1}{|x|\\sqrt{x^2-1}} \\end{array}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169736654830\">Find the derivative of [latex]f(x)=\\tan^{-1} (x^2)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739270394\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739270394\"]\r\n<p id=\"fs-id1169739270394\">Let [latex]g(x)=x^2[\/latex], so [latex]g^{\\prime}(x)=2x[\/latex]. Substituting into [latex]\\frac{d}{dx}(\\tan^{-1} x)=\\large \\frac{1}{1+x^2}[\/latex], we obtain<\/p>\r\n\r\n<div id=\"fs-id1169736661176\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{1}{1+(x^2)^2} \\cdot (2x)[\/latex]<\/div>\r\n<p id=\"fs-id1169736595973\">Simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1169736595976\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{2x}{1+x^4}[\/latex]<\/div>\r\n[\/hidden-answer]<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169739208500\">The position of a particle at time [latex]t[\/latex] is given by [latex]s(t)= \\tan^{-1}\\left(\\dfrac{1}{t}\\right)[\/latex] for [latex]t\\ge \\frac{1}{2}[\/latex]. Find the velocity of the particle at time [latex]t=1[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169736656603\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736656603\"]\r\n<p id=\"fs-id1169736656603\">Begin by differentiating [latex]s(t)[\/latex] in order to find [latex]v(t)[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1169739242485\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=s^{\\prime}(t)=\\dfrac{1}{1+(\\frac{1}{t})^2} \\cdot \\dfrac{-1}{t^2}[\/latex]<\/div>\r\n<p id=\"fs-id1169739282656\">Simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1169736615257\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=-\\dfrac{1}{t^2+1}[\/latex]<\/div>\r\n<p id=\"fs-id1169736594230\" style=\"text-align: center;\">Thus, [latex]v(1)=-\\dfrac{1}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]206679[\/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 differentiation formulas to find derivatives of logarithmic functions and inverse trigonometric functions<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Derivative of the Logarithmic Function<\/h2>\n<p id=\"fs-id1169738222225\">We can use implicit differentiation to find the derivative of the natural logarithmic function by working with its relationship to the natural exponential function.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">derivative of the natural logarithmic function<\/h3>\n<p id=\"fs-id1169737911417\">If [latex]x>0[\/latex] and [latex]y=\\ln x[\/latex], then<\/p>\n<div id=\"fs-id1169738223534\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737765282\">More generally, let [latex]g(x)[\/latex] be a differentiable function. For all values of [latex]x[\/latex] for which [latex]g^{\\prime}(x)>0[\/latex], the derivative of [latex]h(x)=\\ln(g(x))[\/latex] is given by<\/p>\n<div id=\"fs-id1169737919348\" class=\"equation\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{g(x)} g^{\\prime}(x)[\/latex]<\/div>\n<\/section>\n<p>The graph of [latex]y=\\ln x[\/latex] and its derivative [latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex] are shown in Figure 3.<\/p>\n<figure style=\"width: 487px\" 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\/11205536\/CNX_Calc_Figure_03_09_003.jpg\" alt=\"Graph of the function ln x along with its derivative 1\/x. The function ln x is increasing on (0, + \u221e). Its derivative is decreasing but greater than 0 on (0, + \u221e).\" width=\"487\" height=\"209\" \/><figcaption class=\"wp-caption-text\">Figure 3.\u00a0The function [latex]y=\\ln x[\/latex] is increasing on [latex](0,+\\infty)[\/latex]. Its derivative [latex]y^{\\prime} =\\frac{1}{x}[\/latex] is greater than zero on [latex](0,+\\infty)[\/latex].<\/figcaption><\/figure>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738228377\">Find the derivative of [latex]f(x)=\\ln(x^3+3x-4)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738221312\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738221312\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738221312\">Use the derivative of a natural logarithm directly.<\/p>\n<div id=\"fs-id1169738220266\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} f^{\\prime}(x) & =\\frac{1}{x^3+3x-4} \\cdot (3x^2+3) & & & \\text{Use} \\, g(x)=x^3+3x-4 \\, \\text{in} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ & =\\frac{3x^2+3}{x^3+3x-4} & & & \\text{Rewrite.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738106144\">Find the derivative of [latex]f(x)=\\ln\\left(\\dfrac{x^2 \\sin x}{2x+1}\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738219674\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738219674\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738219674\">At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.<\/p>\n<div id=\"fs-id1169738219679\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} f(x) & = \\ln(\\frac{x^2 \\sin x}{2x+1})=2\\ln x+\\ln(\\sin x)-\\ln(2x+1) & & & \\text{Apply properties of logarithms.} \\\\ f^{\\prime}(x) & = \\frac{2}{x} + \\frac{\\cos x}{\\sin x} -\\frac{2}{2x+1} & & & \\text{Apply sum rule and} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ & = \\frac{2}{x} + \\cot x - \\frac{2}{2x+1} & & & \\text{Simplify using the quotient identity for cotangent.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1169738076342\">Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of [latex]y=\\log_b x[\/latex] and [latex]y=b^x[\/latex] for [latex]b>0, \\, b\\ne 1[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">derivatives of general exponential and logarithmic functions<\/h3>\n<p id=\"fs-id1169738186170\">Let [latex]b>0, \\, b\\ne 1[\/latex], and let [latex]g(x)[\/latex] be a differentiable function.<\/p>\n<ol id=\"fs-id1169737998025\">\n<li>If [latex]y=\\log_b x[\/latex], then\n<div id=\"fs-id1169738105090\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x \\ln b}[\/latex]<\/div>\n<p>More generally, if [latex]h(x)=\\log_b (g(x))[\/latex], then for all values of [latex]x[\/latex] for which [latex]g(x)>0[\/latex],<\/p>\n<div id=\"fs-id1169738186308\" class=\"equation\">\n<p style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{g^{\\prime}(x)}{g(x) \\ln b}[\/latex]<\/p>\n<\/div>\n<\/li>\n<li>If [latex]y=b^x[\/latex], then\n<div id=\"fs-id1169738224034\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=b^x \\ln b[\/latex]<\/div>\n<p>More generally, if [latex]h(x)=b^{g(x)}[\/latex], then<\/p>\n<div id=\"fs-id1169738045159\" class=\"equation\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=b^{g(x)} g^{\\prime}(x) \\ln b[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738185259\">Find the derivative of [latex]h(x)= \\dfrac{3^x}{3^x+2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169737700313\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169737700313\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737700313\">Use the quotient rule and the derivative from above.<\/p>\n<div id=\"fs-id1169737700320\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} h^{\\prime}(x) & = \\large \\frac{3^x \\ln 3(3^x+2)-3^x \\ln 3(3^x)}{(3^x+2)^2} & & & \\text{Apply the quotient rule.} \\\\ & = \\large \\frac{2 \\cdot 3^x \\ln 3}{(3^x+2)^2} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738219403\">Find the slope of the line tangent to the graph of [latex]y=\\log_2 (3x+1)[\/latex] at [latex]x=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738045067\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738045067\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738045067\">To find the slope, we must evaluate [latex]\\dfrac{dy}{dx}[\/latex] at [latex]x=1[\/latex]. Using the derivative above, we see that<\/p>\n<div id=\"fs-id1169738197861\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{dy}{dx}=\\dfrac{3}{\\ln 2(3x+1)}[\/latex]<\/div>\n<p id=\"fs-id1169738186987\">By evaluating the derivative at [latex]x=1[\/latex], we see that the tangent line has slope<\/p>\n<div id=\"fs-id1169738187002\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}|_{x=1} =\\frac{3}{4 \\ln 2}=\\frac{3}{\\ln 16}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm288388\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288388&theme=lumen&iframe_resize_id=ohm288388&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<h2>Derivatives of Inverse Trigonometric Functions<\/h2>\n<p id=\"fs-id1169739029363\">The derivatives of inverse trigonometric functions play a crucial role in the study of integration and reveal a fascinating mathematical pattern. Unlike their trigonometric counterparts, the derivatives of inverse trigonometric functions are algebraic rather than trigonometric. This illustrates an important mathematical insight: the derivative of a function does not necessarily share the same type as the original function.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739188589\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\sin^{-1} x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739269789\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739269789\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739269789\">Since for [latex]x[\/latex] in the interval [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}], \\, f(x)= \\sin x[\/latex] is the inverse of [latex]g(x)= \\sin^{-1} x[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex].<\/p>\n<p>Since:<\/p>\n<div id=\"fs-id1169736611678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)= \\cos x[\/latex] and [latex]f^{\\prime}(g(x))= \\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex],<\/div>\n<p id=\"fs-id1169739347090\">we see that:<\/p>\n<div id=\"fs-id1169739369239\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{d}{dx}(\\sin^{-1} x)=\\frac{1}{f^{\\prime}(g(x))}=\\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/div>\n<p><strong>Analysis<\/strong><\/p>\n<p id=\"fs-id1169739111057\">To see that [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex], consider the following argument. Set [latex]\\sin^{-1} x=\\theta[\/latex]. In this case, [latex]\\sin \\theta =x[\/latex] where [latex]-\\frac{\\pi}{2}\\le \\theta \\le \\frac{\\pi}{2}[\/latex]. We begin by considering the case where [latex]0<\\theta <\\frac{\\pi}{2}[\/latex]. Since [latex]\\theta[\/latex] is an acute angle, we may construct a right triangle having acute angle [latex]\\theta[\/latex], a hypotenuse of length 1, and the side opposite angle [latex]\\theta[\/latex] having length [latex]x[\/latex]. From the Pythagorean theorem, the side adjacent to angle [latex]\\theta[\/latex] has length [latex]\\sqrt{1-x^2}[\/latex]. This triangle is shown in Figure 2. Using the triangle, we see that [latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex].<\/p>\n<figure style=\"width: 426px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205436\/CNX_Calc_Figure_03_07_002.jpg\" alt=\"A right triangle with angle \u03b8, opposite side x, hypotenuse 1, and adjacent side equal to the square root of the quantity (1 \u2013 x2).\" width=\"426\" height=\"205\" \/><figcaption class=\"wp-caption-text\">Figure 2. Using a right triangle having acute angle [latex]\\theta[\/latex], a hypotenuse of length 1, and the side opposite angle [latex]\\theta[\/latex] having length [latex]x[\/latex], we can see that [latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex].<\/figcaption><\/figure>\n<div id=\"fs-id1169739303300\" class=\"equation unnumbered\">In the case where [latex]-\\frac{\\pi}{2}<\\theta <0[\/latex], we make the observation that [latex]0<-\\theta<\\frac{\\pi}{2}[\/latex] and hence [latex]\\cos (\\sin^{-1} x)= \\cos \\theta = \\cos (\u2212\\theta )=\\sqrt{1-x^2}[\/latex].<\/div>\n<p id=\"fs-id1169736611289\">Now if [latex]\\theta =\\frac{\\pi}{2}[\/latex] or [latex]\\theta =-\\frac{\\pi}{2}, \\, x=1[\/latex] or [latex]x=-1[\/latex], and since in either case [latex]\\cos \\theta =0[\/latex] and [latex]\\sqrt{1-x^2}=0[\/latex], we have<\/p>\n<div id=\"fs-id1169739340290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex]<\/div>\n<p id=\"fs-id1169739190632\">Consequently, in all cases, [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739190553\">Apply the chain rule to find the derivative of [latex]h(x)=\\sin^{-1} (g(x))[\/latex] and use this result to find the derivative of [latex]h(x)=\\sin^{-1}(2x^3)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739189899\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739189899\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739189899\">Applying the chain rule to [latex]h(x)=\\sin^{-1} (g(x))[\/latex], we have<\/p>\n<div id=\"fs-id1169736596017\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{\\sqrt{1-(g(x))^2}}g^{\\prime}(x)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739269950\">Now let [latex]g(x)=2x^3[\/latex], so [latex]g^{\\prime}(x)=6x^{2}[\/latex]. Substituting into the previous result, we obtain<\/p>\n<div id=\"fs-id1169739336080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} h^{\\prime}(x) & =\\dfrac{1}{\\sqrt{1-4x^6}} \\cdot 6x^{2} \\\\ & =\\dfrac{6x^{2}}{\\sqrt{1-4x^6}} \\end{array}[\/latex]<\/div>\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\/0dlA5QJZYsw?controls=0&amp;start=628&amp;end=723&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\/3.7DerivativesOfInverseFunctions628to723_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.7 Derivatives of Inverse Functions (edited)&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1169739282743\">The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">derivatives of inverse trigonometric functions<\/h3>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\frac{d}{dx}(\\sin^{-1} x)=\\large \\frac{1}{\\sqrt{1-x^2}} & & & & \\frac{d}{dx}(\\cos^{-1} x)=\\large \\frac{-1}{\\sqrt{1-x^2}} \\\\ \\frac{d}{dx}(\\tan^{-1} x)=\\large \\frac{1}{1+x^2} & & & & \\frac{d}{dx}(\\cot^{-1} x)=\\large \\frac{-1}{1+x^2} \\\\ \\frac{d}{dx}(\\sec^{-1} x)=\\large \\frac{1}{|x|\\sqrt{x^2-1}} & & & & \\frac{d}{dx}(\\csc^{-1} x)=\\large \\frac{-1}{|x|\\sqrt{x^2-1}} \\end{array}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169736654830\">Find the derivative of [latex]f(x)=\\tan^{-1} (x^2)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739270394\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739270394\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739270394\">Let [latex]g(x)=x^2[\/latex], so [latex]g^{\\prime}(x)=2x[\/latex]. Substituting into [latex]\\frac{d}{dx}(\\tan^{-1} x)=\\large \\frac{1}{1+x^2}[\/latex], we obtain<\/p>\n<div id=\"fs-id1169736661176\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{1}{1+(x^2)^2} \\cdot (2x)[\/latex]<\/div>\n<p id=\"fs-id1169736595973\">Simplifying, we have<\/p>\n<div id=\"fs-id1169736595976\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{2x}{1+x^4}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739208500\">The position of a particle at time [latex]t[\/latex] is given by [latex]s(t)= \\tan^{-1}\\left(\\dfrac{1}{t}\\right)[\/latex] for [latex]t\\ge \\frac{1}{2}[\/latex]. Find the velocity of the particle at time [latex]t=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169736656603\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169736656603\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736656603\">Begin by differentiating [latex]s(t)[\/latex] in order to find [latex]v(t)[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169739242485\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=s^{\\prime}(t)=\\dfrac{1}{1+(\\frac{1}{t})^2} \\cdot \\dfrac{-1}{t^2}[\/latex]<\/div>\n<p id=\"fs-id1169739282656\">Simplifying, we have<\/p>\n<div id=\"fs-id1169736615257\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=-\\dfrac{1}{t^2+1}[\/latex]<\/div>\n<p id=\"fs-id1169736594230\" style=\"text-align: center;\">Thus, [latex]v(1)=-\\dfrac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm206679\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206679&theme=lumen&iframe_resize_id=ohm206679&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":541,"module-header":"- Select Header -","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/683"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/683\/revisions"}],"predecessor-version":[{"id":2030,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/683\/revisions\/2030"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/541"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/683\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=683"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=683"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=683"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=683"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}