{"id":910,"date":"2025-06-20T17:21:37","date_gmt":"2025-06-20T17:21:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=910"},"modified":"2025-08-28T13:30:35","modified_gmt":"2025-08-28T13:30:35","slug":"sequences-and-series-foundations-background-youll-need-3-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/sequences-and-series-foundations-background-youll-need-3-2\/","title":{"raw":"Power Series and Applications: Background You'll Need 3","rendered":"Power Series and Applications: Background You&#8217;ll Need 3"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Recognize and apply derivative patterns for exponential, logarithmic and trigonometric functions<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivatives of the Sine and Cosine Functions<\/h2>\r\nSimple harmonic motion, a type of periodic motion where the restoring force is directly proportional to the displacement, is best described using trigonometric functions like sine and cosine. The behavior of these functions, particularly how they change over time, is crucial in understanding motion dynamics. The derivatives of sine and cosine functions help us compute velocity and acceleration at any point in the motion, linking theoretical physics closely with calculus.\r\n<p id=\"fs-id1169739274430\">We begin our exploration of the derivative for the sine function by using the limit definition to estimate its derivative.<\/p>\r\n\r\n<section class=\"textbox recall\">For a function [latex]f(x),[\/latex] the derivative [latex]f^{\\prime}(x)[\/latex] is defined as:<center>[latex]f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{f(x+h)-f(x)}{h}[\/latex]<\/center>This allows us to approximate [latex]f^{\\prime}(x)[\/latex] for small values of [latex]h[\/latex] as:\r\n\r\n<center>[latex]f^{\\prime}(x)\\approx \\frac{f(x+h)-f(x)}{h}[\/latex].<\/center><\/section>\r\n<p id=\"fs-id1169738837360\">Using [latex]h=0.01[\/latex], we estimate the derivative of the sine function as follows:<\/p>\r\n\r\n<div id=\"fs-id1169738997799\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\sin x)\\approx \\dfrac{\\sin(x+0.01)-\\sin x}{0.01}[\/latex]<\/div>\r\n<p id=\"fs-id1169739223032\">By defining [latex]D(x)=\\frac{\\sin(x+0.01)-\\sin x}{0.01}[\/latex] and plotting this using a graphing tool, we observe an approximation to the derivative of [latex] \\sin x[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"431\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205417\/CNX_Calc_Figure_03_05_001.jpg\" alt=\"The function D(x) = (sin(x + 0.01) \u2212 sin x)\/0.01 is graphed. It looks a lot like a cosine curve.\" width=\"431\" height=\"392\" \/> Figure 1. The resulting graph of [latex]D(x)[\/latex] closely resembles the cosine curve, which supports the derivative relationship.[\/caption]\r\n<p id=\"fs-id1169739302416\">Upon examination, [latex]D(x)[\/latex] appears to be a close match to the graph of the cosine function. This graphical analysis provides a practical demonstration of the derivative, confirming that the derivative of [latex] \\sin x[\/latex] is indeed [latex] \\cos x[\/latex].<\/p>\r\n<p id=\"fs-id1169738975910\">If we were to follow the same steps to approximate the derivative of the cosine function, we would find that<\/p>\r\n\r\n<div id=\"fs-id1169738962070\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\cos x)=\u2212\\sin x[\/latex]<\/div>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">derivatives of [latex]\\sin x[\/latex] and [latex]\\cos x[\/latex]<\/h3>\r\n<p id=\"fs-id1169738998734\">The derivative of the sine function [latex] \\sin x[\/latex] is the cosine function [latex] \\cos x[\/latex].<\/p>\r\n\r\n<center>[latex]\\frac{d}{dx}(\\sin x)= \\cos x[\/latex]<\/center>The derivative of the cosine function [latex] \\cos x[\/latex] is the negative sine function [latex]\u2212\\sin x[\/latex].\r\n\r\n<center>[latex]\\frac{d}{dx}(\\cos x)=\u2212\\sin x[\/latex]<\/center><\/section>\r\n<p id=\"fs-id1169739186572\">The figure below shows the relationship between the graph of [latex]f(x)= \\sin x[\/latex] and its derivative [latex]f^{\\prime}(x)= \\cos x[\/latex]. Notice that at the points where [latex]f(x)= \\sin x[\/latex] has a horizontal tangent, its derivative [latex]f^{\\prime}(x)= \\cos x[\/latex] takes on the value zero. We also see that where [latex]f(x)= \\sin x[\/latex] is increasing, [latex]f^{\\prime}(x)= \\cos x&gt;0[\/latex] and where [latex]f(x)= \\sin x[\/latex] is decreasing, [latex]f^{\\prime}(x)= \\cos x&lt;0[\/latex].<\/p>\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\/11205423\/CNX_Calc_Figure_03_05_003.jpg\" alt=\"The functions f(x) = sin x and f\u2019(x) = cos x are graphed. It is apparent that when f(x) has a maximum or a minimum that f\u2019(x) = 0.\" width=\"487\" height=\"358\" \/> Figure 3. Where [latex]f(x)[\/latex] has a maximum or a minimum, [latex]f^{\\prime}(x)=0[\/latex]. That is, [latex]f^{\\prime}(x)=0[\/latex] where [latex]f(x)[\/latex] has a horizontal tangent. These points are noted with dots on the graphs.[\/caption]<section class=\"textbox example\">\r\n<p id=\"fs-id1169739269454\">Find the derivative of [latex]f(x)=5x^3 \\sin x[\/latex].<\/p>\r\n[reveal-answer q=\"300277\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"300277\"]\r\n<p id=\"fs-id1169738821957\">Don\u2019t forget to use the product rule.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739028319\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739028319\"]Using the product rule, we have\r\n<div id=\"fs-id1169739001004\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}f^{\\prime}(x) &amp; =\\frac{d}{dx}(5x^3)\\cdot \\sin x+\\frac{d}{dx}(\\sin x)\\cdot 5x^3 \\\\ &amp; =15x^2\\cdot \\sin x+ \\cos x\\cdot 5x^3\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169739036358\">After simplifying, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739269866\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=15x^2 \\sin x+5x^3 \\cos x[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169739229519\">Find the derivative of [latex]g(x)=\\dfrac{\\cos x}{4x^2}[\/latex].<\/p>\r\n[reveal-answer q=\"488399\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"488399\"]\r\n<p id=\"fs-id1169736587923\">Use the quotient rule.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169738969705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738969705\"]\r\n<p id=\"fs-id1169738969705\">By applying the quotient rule, we have<\/p>\r\n\r\n<div id=\"fs-id1169738960497\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{(\u2212\\sin x)4x^2-8x(\\cos x)}{(4x^2)^2}[\/latex].<\/div>\r\n<p id=\"fs-id1169738949562\">Simplifying, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739179211\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}g^{\\prime}(x) &amp; =\\frac{-4x^2 \\sin x-8x \\cos x}{16x^4} \\\\ &amp; =\\frac{\u2212x \\sin x-2 \\cos x}{4x^3} \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]205604[\/ohm_question]<\/section>\r\n<h2>Derivative of the Exponential Function<\/h2>\r\n<p id=\"fs-id1169738221359\">The differentiation of exponential functions [latex]B(x)=b^x,[\/latex] begins by confirming that [latex]b^x[\/latex] is defined for all real numbers and is inherently continuous. We assume [latex]B^{\\prime}(0)[\/latex] exists and is positive. In this context, we delve deeper into proving that [latex]B(x)[\/latex] is differentiable across its entire domain by making one key assumption: there exists a unique value [latex]b &gt; 0[\/latex] for which [latex]B^{\\prime}(0)=1[\/latex].<\/p>\r\n<p id=\"fs-id1169737728646\">The graph of [latex]E(x)=e^x[\/latex] is shown alongside the line [latex]y=x+1[\/latex] in Figure 2, demonstrating that the tangent to [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex] has a slope of [latex]1[\/latex]. This observation supports the hypothesis that the value of [latex]e[\/latex] optimizes the slope at [latex]x=0[\/latex] to exactly [latex]1[\/latex].<\/p>\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\/11205533\/CNX_Calc_Figure_03_09_002.jpg\" alt=\"Graph of the function ex along with its tangent at (0, 1), x + 1.\" width=\"487\" height=\"248\" \/> Figure 2. The tangent line to [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex] has slope 1.[\/caption]\r\n<p id=\"fs-id1169738198749\">Now that we understand the underlying behavior at [latex]x=0[\/latex], let's derive the general derivative formula for [latex]B(x)=b^x, \\, b&gt;0[\/latex]. We start by applying the limit definition of the derivative:<\/p>\r\n\r\n<div id=\"fs-id1169738045860\" class=\"equation\" style=\"text-align: center;\">[latex]B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\dfrac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\dfrac{b^h-1}{h}[\/latex]<\/div>\r\n<p id=\"fs-id1169738187820\">Turning to [latex]B^{\\prime}(x)[\/latex], we obtain the following.<\/p>\r\n\r\n<div id=\"fs-id1169737954073\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} B^{\\prime}(x) &amp; =\\underset{h\\to 0}{\\lim}\\frac{b^{x+h}-b^x}{h} &amp; &amp; &amp; \\text{Apply the limit definition of the derivative.} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{b^xb^h-b^x}{h} &amp; &amp; &amp; \\text{Note that} \\, b^{x+h}=b^x b^h. \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{b^x(b^h-1)}{h} &amp; &amp; &amp; \\text{Factor out} \\, b^x. \\\\ &amp; =b^x\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h} &amp; &amp; &amp; \\text{Apply a property of limits.} \\\\ &amp; =b^x B^{\\prime}(0) &amp; &amp; &amp; \\text{Use} \\, B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\frac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h}. \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169738221160\">We see that on the basis of the assumption that [latex]B(x)=b^x[\/latex] is differentiable at [latex]0, \\, B(x)[\/latex] is not only differentiable everywhere, but its derivative is<\/p>\r\n\r\n<div id=\"fs-id1169738221385\" class=\"equation\" style=\"text-align: center;\">[latex]B^{\\prime}(x)=b^x B^{\\prime}(0)[\/latex]<\/div>\r\nFor [latex]E(x)=e^x, \\, E^{\\prime}(0)=1[\/latex]. Thus, we have [latex]E^{\\prime}(x)=e^x[\/latex]. (The value of [latex]B^{\\prime}(0)[\/latex] for an arbitrary function of the form [latex]B(x)=b^x, \\, b&gt;0[\/latex], will be derived later.)\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">derivative of the natural exponential function<\/h3>\r\n<p id=\"fs-id1169738220224\">Let [latex]E(x)=e^x[\/latex] be the natural exponential function. Then<\/p>\r\n\r\n<div id=\"fs-id1169737928243\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]E^{\\prime}(x)=e^x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737142141\">In general,<\/p>\r\n\r\n<div id=\"fs-id1169738124964\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\\prime}(x)[\/latex]<\/div>\r\nIf it helps, think of the formula as the chain rule being applied to natural exponential functions. The derivative of [latex]{e}[\/latex] raised to the power of a function will simply be\u00a0[latex]{e}[\/latex] raised to the power of the function multiplied by the derivative of that function.\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169738187154\">Find the derivative of [latex]f(x)=e^{\\tan (2x)}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169737140844\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737140844\"]\r\n<p id=\"fs-id1169737140844\">Using the derivative formula and the chain rule,<\/p>\r\n\r\n<div id=\"fs-id1169738048872\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime}(x) &amp; =e^{\\tan (2x)}\\frac{d}{dx}(\\tan (2x)) \\\\ &amp; = e^{\\tan (2x)} \\sec^2 (2x) \\cdot 2. \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">\r\n<p id=\"fs-id1169737766547\">Find the derivative of [latex]y=\\dfrac{e^{x^2}}{x}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169737928258\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737928258\"]\r\n<p id=\"fs-id1169737928258\">Use the derivative of the natural exponential function, the quotient rule, and the chain rule.<\/p>\r\n\r\n<div id=\"fs-id1169737928262\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} y^{\\prime} &amp; =\\large \\frac{(e^{x^2} \\cdot 2x) \\cdot x - 1 \\cdot e^{x^2}}{x^2} &amp; &amp; &amp; \\text{Apply the quotient rule.} \\\\ &amp; = \\large \\frac{e^{x^2}(2x^2-1)}{x^2} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]33753[\/ohm_question]<\/section>\r\n<h2>Derivative of the Logarithmic Function<\/h2>\r\n<p id=\"fs-id1169738222225\">Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic 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><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]\r\n\r\n<\/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]\r\n\r\n<\/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]\r\n\r\n<\/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]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]288388[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Recognize and apply derivative patterns for exponential, logarithmic and trigonometric functions<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives of the Sine and Cosine Functions<\/h2>\n<p>Simple harmonic motion, a type of periodic motion where the restoring force is directly proportional to the displacement, is best described using trigonometric functions like sine and cosine. The behavior of these functions, particularly how they change over time, is crucial in understanding motion dynamics. The derivatives of sine and cosine functions help us compute velocity and acceleration at any point in the motion, linking theoretical physics closely with calculus.<\/p>\n<p id=\"fs-id1169739274430\">We begin our exploration of the derivative for the sine function by using the limit definition to estimate its derivative.<\/p>\n<section class=\"textbox recall\">For a function [latex]f(x),[\/latex] the derivative [latex]f^{\\prime}(x)[\/latex] is defined as:<\/p>\n<div style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{f(x+h)-f(x)}{h}[\/latex]<\/div>\n<p>This allows us to approximate [latex]f^{\\prime}(x)[\/latex] for small values of [latex]h[\/latex] as:<\/p>\n<div style=\"text-align: center;\">[latex]f^{\\prime}(x)\\approx \\frac{f(x+h)-f(x)}{h}[\/latex].<\/div>\n<\/section>\n<p id=\"fs-id1169738837360\">Using [latex]h=0.01[\/latex], we estimate the derivative of the sine function as follows:<\/p>\n<div id=\"fs-id1169738997799\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\sin x)\\approx \\dfrac{\\sin(x+0.01)-\\sin x}{0.01}[\/latex]<\/div>\n<p id=\"fs-id1169739223032\">By defining [latex]D(x)=\\frac{\\sin(x+0.01)-\\sin x}{0.01}[\/latex] and plotting this using a graphing tool, we observe an approximation to the derivative of [latex]\\sin x[\/latex].<\/p>\n<figure style=\"width: 431px\" 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\/11205417\/CNX_Calc_Figure_03_05_001.jpg\" alt=\"The function D(x) = (sin(x + 0.01) \u2212 sin x)\/0.01 is graphed. It looks a lot like a cosine curve.\" width=\"431\" height=\"392\" \/><figcaption class=\"wp-caption-text\">Figure 1. The resulting graph of [latex]D(x)[\/latex] closely resembles the cosine curve, which supports the derivative relationship.<\/figcaption><\/figure>\n<p id=\"fs-id1169739302416\">Upon examination, [latex]D(x)[\/latex] appears to be a close match to the graph of the cosine function. This graphical analysis provides a practical demonstration of the derivative, confirming that the derivative of [latex]\\sin x[\/latex] is indeed [latex]\\cos x[\/latex].<\/p>\n<p id=\"fs-id1169738975910\">If we were to follow the same steps to approximate the derivative of the cosine function, we would find that<\/p>\n<div id=\"fs-id1169738962070\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\cos x)=\u2212\\sin x[\/latex]<\/div>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">derivatives of [latex]\\sin x[\/latex] and [latex]\\cos x[\/latex]<\/h3>\n<p id=\"fs-id1169738998734\">The derivative of the sine function [latex]\\sin x[\/latex] is the cosine function [latex]\\cos x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\sin x)= \\cos x[\/latex]<\/div>\n<p>The derivative of the cosine function [latex]\\cos x[\/latex] is the negative sine function [latex]\u2212\\sin x[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\cos x)=\u2212\\sin x[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1169739186572\">The figure below shows the relationship between the graph of [latex]f(x)= \\sin x[\/latex] and its derivative [latex]f^{\\prime}(x)= \\cos x[\/latex]. Notice that at the points where [latex]f(x)= \\sin x[\/latex] has a horizontal tangent, its derivative [latex]f^{\\prime}(x)= \\cos x[\/latex] takes on the value zero. We also see that where [latex]f(x)= \\sin x[\/latex] is increasing, [latex]f^{\\prime}(x)= \\cos x>0[\/latex] and where [latex]f(x)= \\sin x[\/latex] is decreasing, [latex]f^{\\prime}(x)= \\cos x<0[\/latex].<\/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\/11205423\/CNX_Calc_Figure_03_05_003.jpg\" alt=\"The functions f(x) = sin x and f\u2019(x) = cos x are graphed. It is apparent that when f(x) has a maximum or a minimum that f\u2019(x) = 0.\" width=\"487\" height=\"358\" \/><figcaption class=\"wp-caption-text\">Figure 3. Where [latex]f(x)[\/latex] has a maximum or a minimum, [latex]f^{\\prime}(x)=0[\/latex]. That is, [latex]f^{\\prime}(x)=0[\/latex] where [latex]f(x)[\/latex] has a horizontal tangent. These points are noted with dots on the graphs.<\/figcaption><\/figure>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739269454\">Find the derivative of [latex]f(x)=5x^3 \\sin x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q300277\">Hint<\/button><\/p>\n<div id=\"q300277\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738821957\">Don\u2019t forget to use the product rule.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739028319\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739028319\" class=\"hidden-answer\" style=\"display: none\">Using the product rule, we have<\/p>\n<div id=\"fs-id1169739001004\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}f^{\\prime}(x) & =\\frac{d}{dx}(5x^3)\\cdot \\sin x+\\frac{d}{dx}(\\sin x)\\cdot 5x^3 \\\\ & =15x^2\\cdot \\sin x+ \\cos x\\cdot 5x^3\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169739036358\">After simplifying, we obtain<\/p>\n<div id=\"fs-id1169739269866\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=15x^2 \\sin x+5x^3 \\cos x[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739229519\">Find the derivative of [latex]g(x)=\\dfrac{\\cos x}{4x^2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q488399\">Hint<\/button><\/p>\n<div id=\"q488399\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736587923\">Use the quotient rule.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738969705\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738969705\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738969705\">By applying the quotient rule, we have<\/p>\n<div id=\"fs-id1169738960497\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{(\u2212\\sin x)4x^2-8x(\\cos x)}{(4x^2)^2}[\/latex].<\/div>\n<p id=\"fs-id1169738949562\">Simplifying, we obtain<\/p>\n<div id=\"fs-id1169739179211\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}g^{\\prime}(x) & =\\frac{-4x^2 \\sin x-8x \\cos x}{16x^4} \\\\ & =\\frac{\u2212x \\sin x-2 \\cos x}{4x^3} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm205604\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=205604&theme=lumen&iframe_resize_id=ohm205604&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h2>Derivative of the Exponential Function<\/h2>\n<p id=\"fs-id1169738221359\">The differentiation of exponential functions [latex]B(x)=b^x,[\/latex] begins by confirming that [latex]b^x[\/latex] is defined for all real numbers and is inherently continuous. We assume [latex]B^{\\prime}(0)[\/latex] exists and is positive. In this context, we delve deeper into proving that [latex]B(x)[\/latex] is differentiable across its entire domain by making one key assumption: there exists a unique value [latex]b > 0[\/latex] for which [latex]B^{\\prime}(0)=1[\/latex].<\/p>\n<p id=\"fs-id1169737728646\">The graph of [latex]E(x)=e^x[\/latex] is shown alongside the line [latex]y=x+1[\/latex] in Figure 2, demonstrating that the tangent to [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex] has a slope of [latex]1[\/latex]. This observation supports the hypothesis that the value of [latex]e[\/latex] optimizes the slope at [latex]x=0[\/latex] to exactly [latex]1[\/latex].<\/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\/11205533\/CNX_Calc_Figure_03_09_002.jpg\" alt=\"Graph of the function ex along with its tangent at (0, 1), x + 1.\" width=\"487\" height=\"248\" \/><figcaption class=\"wp-caption-text\">Figure 2. The tangent line to [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex] has slope 1.<\/figcaption><\/figure>\n<p id=\"fs-id1169738198749\">Now that we understand the underlying behavior at [latex]x=0[\/latex], let&#8217;s derive the general derivative formula for [latex]B(x)=b^x, \\, b>0[\/latex]. We start by applying the limit definition of the derivative:<\/p>\n<div id=\"fs-id1169738045860\" class=\"equation\" style=\"text-align: center;\">[latex]B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\dfrac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\dfrac{b^h-1}{h}[\/latex]<\/div>\n<p id=\"fs-id1169738187820\">Turning to [latex]B^{\\prime}(x)[\/latex], we obtain the following.<\/p>\n<div id=\"fs-id1169737954073\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} B^{\\prime}(x) & =\\underset{h\\to 0}{\\lim}\\frac{b^{x+h}-b^x}{h} & & & \\text{Apply the limit definition of the derivative.} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{b^xb^h-b^x}{h} & & & \\text{Note that} \\, b^{x+h}=b^x b^h. \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{b^x(b^h-1)}{h} & & & \\text{Factor out} \\, b^x. \\\\ & =b^x\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h} & & & \\text{Apply a property of limits.} \\\\ & =b^x B^{\\prime}(0) & & & \\text{Use} \\, B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\frac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h}. \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169738221160\">We see that on the basis of the assumption that [latex]B(x)=b^x[\/latex] is differentiable at [latex]0, \\, B(x)[\/latex] is not only differentiable everywhere, but its derivative is<\/p>\n<div id=\"fs-id1169738221385\" class=\"equation\" style=\"text-align: center;\">[latex]B^{\\prime}(x)=b^x B^{\\prime}(0)[\/latex]<\/div>\n<p>For [latex]E(x)=e^x, \\, E^{\\prime}(0)=1[\/latex]. Thus, we have [latex]E^{\\prime}(x)=e^x[\/latex]. (The value of [latex]B^{\\prime}(0)[\/latex] for an arbitrary function of the form [latex]B(x)=b^x, \\, b>0[\/latex], will be derived later.)<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">derivative of the natural exponential function<\/h3>\n<p id=\"fs-id1169738220224\">Let [latex]E(x)=e^x[\/latex] be the natural exponential function. Then<\/p>\n<div id=\"fs-id1169737928243\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]E^{\\prime}(x)=e^x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737142141\">In general,<\/p>\n<div id=\"fs-id1169738124964\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\\prime}(x)[\/latex]<\/div>\n<p>If it helps, think of the formula as the chain rule being applied to natural exponential functions. The derivative of [latex]{e}[\/latex] raised to the power of a function will simply be\u00a0[latex]{e}[\/latex] raised to the power of the function multiplied by the derivative of that function.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738187154\">Find the derivative of [latex]f(x)=e^{\\tan (2x)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169737140844\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169737140844\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737140844\">Using the derivative formula and the chain rule,<\/p>\n<div id=\"fs-id1169738048872\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime}(x) & =e^{\\tan (2x)}\\frac{d}{dx}(\\tan (2x)) \\\\ & = e^{\\tan (2x)} \\sec^2 (2x) \\cdot 2. \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169737766547\">Find the derivative of [latex]y=\\dfrac{e^{x^2}}{x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169737928258\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169737928258\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737928258\">Use the derivative of the natural exponential function, the quotient rule, and the chain rule.<\/p>\n<div id=\"fs-id1169737928262\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} y^{\\prime} & =\\large \\frac{(e^{x^2} \\cdot 2x) \\cdot x - 1 \\cdot e^{x^2}}{x^2} & & & \\text{Apply the quotient rule.} \\\\ & = \\large \\frac{e^{x^2}(2x^2-1)}{x^2} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm33753\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33753&theme=lumen&iframe_resize_id=ohm33753&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h2>Derivative of the Logarithmic Function<\/h2>\n<p id=\"fs-id1169738222225\">Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic 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<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><\/section>\n","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":673,"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\/910"}],"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":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/910\/revisions"}],"predecessor-version":[{"id":2056,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/910\/revisions\/2056"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/673"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/910\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=910"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=910"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=910"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=910"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}