{"id":264,"date":"2023-09-20T22:48:40","date_gmt":"2023-09-20T22:48:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivative-of-the-exponential-function\/"},"modified":"2024-08-05T12:57:00","modified_gmt":"2024-08-05T12:57:00","slug":"derivatives-of-exponential-and-logarithmic-functions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-of-exponential-and-logarithmic-functions-learn-it-1\/","title":{"raw":"Derivatives of Exponential and Logarithmic Functions: Learn It 1","rendered":"Derivatives of Exponential and Logarithmic Functions: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Determine the derivatives of exponential and logarithmic functions<\/li>\r\n\t<li>Apply logarithmic differentiation to find derivatives<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p>Exponential and logarithmic functions are crucial in modeling growth processes, such as population dynamics and radioactive decay. These functions also simplify complex mathematical expressions. In this section, we delve deeper into their derivatives, highlighting the uniqueness of these functions, particularly through the properties of the natural exponential function, [latex]e^x[\/latex].<\/p>\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] where [latex]b&gt;0[\/latex], begins with recognizing that these functions are defined for all real numbers and are continuous. This universality stems from the properties defined for exponentiation over rational and irrational numbers alike.<\/p>\r\n<p>Historically, exponentiation for integers was defined straightforwardly\u2014for example, [latex]b^n[\/latex] where [latex]n[\/latex] is an integer. For non-integer exponents, [latex]b^r[\/latex] (where [latex]r[\/latex] is any real number), we interpret this through a limiting process. Specifically, [latex]b^r[\/latex] is understood as the limit of [latex]b^x[\/latex] as [latex]x[\/latex] approaches [latex]r[\/latex] from rational values, ensuring the function\u2019s continuity across all real numbers.<\/p>\r\n<p>To illustrate, consider the function [latex]{4}^{x}[\/latex] evaluated at values close to [latex]\\pi[\/latex]. We may view [latex]{4}^{\\pi}[\/latex] as the number satisfying:<\/p>\r\n<div id=\"fs-id1169738212703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}4^3&lt;4^{\\pi}&lt;4^4, \\\\ 4^{3.1}&lt;4^{\\pi}&lt;4^{3.2}, \\\\ 4^{3.14}&lt;4^{\\pi}&lt;4^{3.15},\\\\ 4^{3.141}&lt;4^{\\pi}&lt;4^{3.142}, \\\\ 4^{3.1415}&lt;4^{\\pi}&lt;4^{3.1416}, \\\\ \\cdots \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169737765653\">As we see in the following table, [latex]4^{\\pi}\\approx 77.88[\/latex].<\/p>\r\n<table id=\"fs-id1169738222068\" summary=\"This table has seven rows and four columns. The first row is a header row and it labels each column. The first column header is x, the second column header is 4x, the third column header is x, and the fourth column header is 4x. Under the first column are the values 43, 43.1, 43.14, 43.141, 43.1415. Under the second column are the values 64, 73.5166947198, 77.7084726013, 77.8162741237, 77.8702309526, 77.8799471543. Under the third column are the values 43.141593, 43.1416, 43.142, 43.15, 43.2, and 44. Under the fourth column are the values 77.8802710486, 77.8810268071, 77.9242251944, 78.7932424541, 84.4485062895, and 256.\">\r\n<caption>Approximating a Value of [latex]4^{\\pi}[\/latex]<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]4^x[\/latex]<\/th>\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]4^x[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]4^3[\/latex]<\/td>\r\n<td>[latex]64[\/latex]<\/td>\r\n<td>[latex]4^{3.141593}[\/latex]<\/td>\r\n<td>[latex]77.8802710486[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.1}[\/latex]<\/td>\r\n<td>[latex]73.5166947198[\/latex]<\/td>\r\n<td>[latex]4^{3.1416}[\/latex]<\/td>\r\n<td>[latex]77.8810268071[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.14}[\/latex]<\/td>\r\n<td>[latex]77.7084726013[\/latex]<\/td>\r\n<td>[latex]4^{3.142}[\/latex]<\/td>\r\n<td>[latex]77.9242251944[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.141}[\/latex]<\/td>\r\n<td>[latex]77.8162741237[\/latex]<\/td>\r\n<td>[latex]4^{3.15}[\/latex]<\/td>\r\n<td>[latex]78.7932424541[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.1415}[\/latex]<\/td>\r\n<td>[latex]77.8702309526[\/latex]<\/td>\r\n<td>[latex]4^{3.2}[\/latex]<\/td>\r\n<td>[latex]84.4485062895[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.14159}[\/latex]<\/td>\r\n<td>[latex]77.8799471543[\/latex]<\/td>\r\n<td>[latex]4^4[\/latex]<\/td>\r\n<td>[latex]256[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>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]. This value, defined as [latex]e[\/latex], demonstrates unique properties, marking it as the base of natural logarithms.<\/p>\r\n<p>Figure 1 provides graphs of the functions [latex]y=2^x, \\, y=3^x, \\, y=2.7^x[\/latex], and [latex]y=2.8^x[\/latex]. Observing the slopes of tangent lines at [latex]x=0[\/latex] for these functions provides empirical evidence suggesting that [latex]e[\/latex] is approximately between [latex]2.7[\/latex] and [latex]2.8[\/latex]. This observation is supported by the graph where the tangent line at [latex]x=0[\/latex] for [latex]y=e^x[\/latex] has a slope of [latex]1[\/latex], uniquely defining [latex]e[\/latex] as approximately [latex]2.718[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205529\/CNX_Calc_Figure_03_09_001.jpg\" alt=\"The graphs of 3x, 2.8x, 2.7x, and 2x are shown. In quadrant I, their order from least to greatest is 2x, 2.7x, 2.8x, and 3x. In quadrant II, this order is reversed. All cross the y-axis at (0, 1).\" width=\"731\" height=\"592\" \/> Figure 1. The graph of [latex]E(x)=e^x[\/latex] is between [latex]y=2^x[\/latex] and [latex]y=3^x[\/latex].[\/caption]\r\n\r\n<p id=\"fs-id1169737728646\">The <strong>natural exponential function, <\/strong>[latex]E(x)=e^x[\/latex], and its inverse, the <strong>natural logarithm\u00a0<\/strong>[latex]L(x)=\\log_e x=\\ln x[\/latex] are essential for understanding continuous growth and decay processes modeled in various scientific and financial contexts.<\/p>\r\n<p>To estimate the value of [latex]e[\/latex], we consider the derivative of exponential functions of the form [latex]B(x)=b^x[\/latex] at [latex]x=0[\/latex], denoted as [latex]B^{\\prime}(0)[\/latex].<\/p>\r\n<p>To accurately determine [latex]B^{\\prime}(0)[\/latex], we select values of x very close to zero, such as [latex]x=0.00001[\/latex] and [latex]x=-0.00001[\/latex]. This method allows us to find [latex]B^{\\prime}(0)[\/latex] and compare it against our hypothesized value of [latex]b[\/latex] that satisfies [latex]B^{\\prime}(0)=1[\/latex], indicating [latex]b=e[\/latex].<\/p>\r\n<p id=\"fs-id1169737952624\">The table below illustrates how varying [latex]b[\/latex] around the theoretical value of [latex]e[\/latex] allows us to pinpoint the exact value where [latex]B^{\\prime}(0) \\approx 1[\/latex] , suggesting [latex]e \\approx 2.718[\/latex].<\/p>\r\n<table id=\"fs-id1169738019199\" summary=\"This table has six rows and four columns. The first row is a header row and it labels each column. The first column header is b, the second column header is (b\u22120.00001 \u2212 1)\/\u22120.00001 &lt; B\u2019(0) &lt; (b0.00001 \u2212 1)\/0.00001, the third column header is b, and the fourth column header is (b\u22120.00001 \u2212 1)\/\u22120.00001 &lt; B\u2019(0) &lt; (b0.00001 \u2212 1)\/0.00001. Under the first column are the values 2, 2.7, 2.71, 2.718, and 2.7182. Under the second column are the values 0.693145&lt;B\u2019(0)&lt;0.69315, 0.993247&lt;B\u2019(0)&lt; 0.993257, 0.996944&lt;B\u2019(0)&lt;0.996954, 0.999891&lt;B\u2019(0)&lt; 0.999901, and 0.999965&lt;B\u2019(0)&lt;0.999975. Under the third column are the values 2.7183, 2.719, 2.72, 2.8, and 3. Under the fourth column are the values 1.000002&lt;B\u2019(0)&lt; 1.000012, 1.000259&lt;B\u2019(0)&lt; 1.000269, 1.000627&lt;B\u2019(0)&lt;1.000637, 1.029614&lt;B\u2019(0)&lt;1.029625, and 1.098606&lt;B\u2019(00&lt;1.098618.\">\r\n<caption>Estimating a Value of [latex]e[\/latex]<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]b[\/latex]<\/th>\r\n<th>[latex]\\frac{b^{-0.00001}-1}{-0.00001} &lt; B^{\\prime}(0) &lt; \\frac{b^{0.00001}-1}{0.00001}[\/latex]<\/th>\r\n<th>[latex]b[\/latex]<\/th>\r\n<th>[latex]\\frac{b^{-0.00001}-1}{-0.00001} &lt; B^{\\prime}(0) &lt; \\frac{b^{0.00001}-1}{0.00001}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]0.693145 &lt; B^{\\prime}(0) &lt; 0.69315[\/latex]<\/td>\r\n<td>[latex]2.7183[\/latex]<\/td>\r\n<td>[latex]1.000002 &lt; B^{\\prime}(0) &lt; 1.000012[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]2.7[\/latex]<\/td>\r\n<td>[latex]0.993247 &lt; B^{\\prime}(0) &lt; 0.993257[\/latex]<\/td>\r\n<td>[latex]2.719[\/latex]<\/td>\r\n<td>[latex]1.000259 &lt; B^{\\prime}(0) &lt; 1.000269[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]2.71[\/latex]<\/td>\r\n<td>[latex]0.996944 &lt; B^{\\prime}(0) &lt; 0.996954[\/latex]<\/td>\r\n<td>[latex]2.72[\/latex]<\/td>\r\n<td>[latex]1.000627 &lt; B^{\\prime}(0) &lt; 1.000637[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]2.718[\/latex]<\/td>\r\n<td>[latex]0.999891 &lt; B^{\\prime}(0) &lt; 0.999901[\/latex]<\/td>\r\n<td>[latex]2.8[\/latex]<\/td>\r\n<td>[latex]1.029614 &lt; B^{\\prime}(0) &lt; 1.029625[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]2.7182[\/latex]<\/td>\r\n<td>[latex]0.999965 &lt; B^{\\prime}(0) &lt; 0.999975[\/latex]<\/td>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]1.098606 &lt; B^{\\prime}(0) &lt; 1.098618[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1169737978597\">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\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<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<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<div id=\"fs-id1169738221385\" class=\"equation\" style=\"text-align: center;\">[latex]B^{\\prime}(x)=b^x B^{\\prime}(0)[\/latex]<\/div>\r\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&gt;0[\/latex], will be derived later.)<\/p>\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<div id=\"fs-id1169737928243\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]E^{\\prime}(x)=e^x[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169737142141\">In general,<\/p>\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\n\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.<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169738187154\">Find the derivative of [latex]f(x)=e^{\\tan (2x)}[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169737140844\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169737140844\"]<\/p>\r\n<p id=\"fs-id1169737140844\">Using the derivative formula and the chain rule,<\/p>\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\r\n[\/hidden-answer]<\/section>\r\n<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<p>[reveal-answer q=\"fs-id1169737928258\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169737928258\"]<\/p>\r\n<p id=\"fs-id1169737928258\">Use the derivative of the natural exponential function, the quotient rule, and the chain rule.<\/p>\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<p>&nbsp;<\/p>\r\n\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]33753[\/ohm_question]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169737151943\">A colony of mosquitoes has an initial population of [latex]1000.[\/latex] After [latex]t[\/latex] days, the population is given by [latex]A(t)=1000e^{0.3t}[\/latex]. Show that the ratio of the rate of change of the population, [latex]A^{\\prime}(t)[\/latex], to the population size, [latex]A(t)[\/latex] is constant.<\/p>\r\n<p>[reveal-answer q=\"fs-id1169738212487\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169738212487\"]<\/p>\r\n<p id=\"fs-id1169738212487\">First find [latex]A^{\\prime}(t)[\/latex]. By using the chain rule, we have [latex]A^{\\prime}(t)=300e^{0.3t}[\/latex]. Thus, the ratio of the rate of change of the population to the population size is given by<\/p>\r\n<div id=\"fs-id1169738227689\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\large \\frac{A^{\\prime}(t)}{A(t)} \\normalsize = \\large \\frac{300e^{0.3t}}{1000e^{0.3t}}=0.3[\/latex]<\/div>\r\n<p id=\"fs-id1169737145031\">The ratio of the rate of change of the population to the population size is the constant [latex]0.3[\/latex].<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=163&amp;end=240&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.9DerivativesOfExponentialAndLogarithmic163to240_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.9 Derivatives of Exponential and Logarithmic Functions\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Determine the derivatives of exponential and logarithmic functions<\/li>\n<li>Apply logarithmic differentiation to find derivatives<\/li>\n<\/ul>\n<\/section>\n<p>Exponential and logarithmic functions are crucial in modeling growth processes, such as population dynamics and radioactive decay. These functions also simplify complex mathematical expressions. In this section, we delve deeper into their derivatives, highlighting the uniqueness of these functions, particularly through the properties of the natural exponential function, [latex]e^x[\/latex].<\/p>\n<h2>Derivative of the Exponential Function<\/h2>\n<p id=\"fs-id1169738221359\">The differentiation of exponential functions [latex]B(x)=b^x[\/latex] where [latex]b>0[\/latex], begins with recognizing that these functions are defined for all real numbers and are continuous. This universality stems from the properties defined for exponentiation over rational and irrational numbers alike.<\/p>\n<p>Historically, exponentiation for integers was defined straightforwardly\u2014for example, [latex]b^n[\/latex] where [latex]n[\/latex] is an integer. For non-integer exponents, [latex]b^r[\/latex] (where [latex]r[\/latex] is any real number), we interpret this through a limiting process. Specifically, [latex]b^r[\/latex] is understood as the limit of [latex]b^x[\/latex] as [latex]x[\/latex] approaches [latex]r[\/latex] from rational values, ensuring the function\u2019s continuity across all real numbers.<\/p>\n<p>To illustrate, consider the function [latex]{4}^{x}[\/latex] evaluated at values close to [latex]\\pi[\/latex]. We may view [latex]{4}^{\\pi}[\/latex] as the number satisfying:<\/p>\n<div id=\"fs-id1169738212703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}4^3<4^{\\pi}<4^4, \\\\ 4^{3.1}<4^{\\pi}<4^{3.2}, \\\\ 4^{3.14}<4^{\\pi}<4^{3.15},\\\\ 4^{3.141}<4^{\\pi}<4^{3.142}, \\\\ 4^{3.1415}<4^{\\pi}<4^{3.1416}, \\\\ \\cdots \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169737765653\">As we see in the following table, [latex]4^{\\pi}\\approx 77.88[\/latex].<\/p>\n<table id=\"fs-id1169738222068\" summary=\"This table has seven rows and four columns. The first row is a header row and it labels each column. The first column header is x, the second column header is 4x, the third column header is x, and the fourth column header is 4x. Under the first column are the values 43, 43.1, 43.14, 43.141, 43.1415. Under the second column are the values 64, 73.5166947198, 77.7084726013, 77.8162741237, 77.8702309526, 77.8799471543. Under the third column are the values 43.141593, 43.1416, 43.142, 43.15, 43.2, and 44. Under the fourth column are the values 77.8802710486, 77.8810268071, 77.9242251944, 78.7932424541, 84.4485062895, and 256.\">\n<caption>Approximating a Value of [latex]4^{\\pi}[\/latex]<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]4^x[\/latex]<\/th>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]4^x[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]4^3[\/latex]<\/td>\n<td>[latex]64[\/latex]<\/td>\n<td>[latex]4^{3.141593}[\/latex]<\/td>\n<td>[latex]77.8802710486[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.1}[\/latex]<\/td>\n<td>[latex]73.5166947198[\/latex]<\/td>\n<td>[latex]4^{3.1416}[\/latex]<\/td>\n<td>[latex]77.8810268071[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.14}[\/latex]<\/td>\n<td>[latex]77.7084726013[\/latex]<\/td>\n<td>[latex]4^{3.142}[\/latex]<\/td>\n<td>[latex]77.9242251944[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.141}[\/latex]<\/td>\n<td>[latex]77.8162741237[\/latex]<\/td>\n<td>[latex]4^{3.15}[\/latex]<\/td>\n<td>[latex]78.7932424541[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.1415}[\/latex]<\/td>\n<td>[latex]77.8702309526[\/latex]<\/td>\n<td>[latex]4^{3.2}[\/latex]<\/td>\n<td>[latex]84.4485062895[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.14159}[\/latex]<\/td>\n<td>[latex]77.8799471543[\/latex]<\/td>\n<td>[latex]4^4[\/latex]<\/td>\n<td>[latex]256[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>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]. This value, defined as [latex]e[\/latex], demonstrates unique properties, marking it as the base of natural logarithms.<\/p>\n<p>Figure 1 provides graphs of the functions [latex]y=2^x, \\, y=3^x, \\, y=2.7^x[\/latex], and [latex]y=2.8^x[\/latex]. Observing the slopes of tangent lines at [latex]x=0[\/latex] for these functions provides empirical evidence suggesting that [latex]e[\/latex] is approximately between [latex]2.7[\/latex] and [latex]2.8[\/latex]. This observation is supported by the graph where the tangent line at [latex]x=0[\/latex] for [latex]y=e^x[\/latex] has a slope of [latex]1[\/latex], uniquely defining [latex]e[\/latex] as approximately [latex]2.718[\/latex].<\/p>\n<figure style=\"width: 731px\" 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\/11205529\/CNX_Calc_Figure_03_09_001.jpg\" alt=\"The graphs of 3x, 2.8x, 2.7x, and 2x are shown. In quadrant I, their order from least to greatest is 2x, 2.7x, 2.8x, and 3x. In quadrant II, this order is reversed. All cross the y-axis at (0, 1).\" width=\"731\" height=\"592\" \/><figcaption class=\"wp-caption-text\">Figure 1. The graph of [latex]E(x)=e^x[\/latex] is between [latex]y=2^x[\/latex] and [latex]y=3^x[\/latex].<\/figcaption><\/figure>\n<p id=\"fs-id1169737728646\">The <strong>natural exponential function, <\/strong>[latex]E(x)=e^x[\/latex], and its inverse, the <strong>natural logarithm\u00a0<\/strong>[latex]L(x)=\\log_e x=\\ln x[\/latex] are essential for understanding continuous growth and decay processes modeled in various scientific and financial contexts.<\/p>\n<p>To estimate the value of [latex]e[\/latex], we consider the derivative of exponential functions of the form [latex]B(x)=b^x[\/latex] at [latex]x=0[\/latex], denoted as [latex]B^{\\prime}(0)[\/latex].<\/p>\n<p>To accurately determine [latex]B^{\\prime}(0)[\/latex], we select values of x very close to zero, such as [latex]x=0.00001[\/latex] and [latex]x=-0.00001[\/latex]. This method allows us to find [latex]B^{\\prime}(0)[\/latex] and compare it against our hypothesized value of [latex]b[\/latex] that satisfies [latex]B^{\\prime}(0)=1[\/latex], indicating [latex]b=e[\/latex].<\/p>\n<p id=\"fs-id1169737952624\">The table below illustrates how varying [latex]b[\/latex] around the theoretical value of [latex]e[\/latex] allows us to pinpoint the exact value where [latex]B^{\\prime}(0) \\approx 1[\/latex] , suggesting [latex]e \\approx 2.718[\/latex].<\/p>\n<table id=\"fs-id1169738019199\" summary=\"This table has six rows and four columns. The first row is a header row and it labels each column. The first column header is b, the second column header is (b\u22120.00001 \u2212 1)\/\u22120.00001 &lt; B\u2019(0) &lt; (b0.00001 \u2212 1)\/0.00001, the third column header is b, and the fourth column header is (b\u22120.00001 \u2212 1)\/\u22120.00001 &lt; B\u2019(0) &lt; (b0.00001 \u2212 1)\/0.00001. Under the first column are the values 2, 2.7, 2.71, 2.718, and 2.7182. Under the second column are the values 0.693145&lt;B\u2019(0)&lt;0.69315, 0.993247&lt;B\u2019(0)&lt; 0.993257, 0.996944&lt;B\u2019(0)&lt;0.996954, 0.999891&lt;B\u2019(0)&lt; 0.999901, and 0.999965&lt;B\u2019(0)&lt;0.999975. Under the third column are the values 2.7183, 2.719, 2.72, 2.8, and 3. Under the fourth column are the values 1.000002&lt;B\u2019(0)&lt; 1.000012, 1.000259&lt;B\u2019(0)&lt; 1.000269, 1.000627&lt;B\u2019(0)&lt;1.000637, 1.029614&lt;B\u2019(0)&lt;1.029625, and 1.098606&lt;B\u2019(00&lt;1.098618.\">\n<caption>Estimating a Value of [latex]e[\/latex]<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>[latex]b[\/latex]<\/th>\n<th>[latex]\\frac{b^{-0.00001}-1}{-0.00001} < B^{\\prime}(0) < \\frac{b^{0.00001}-1}{0.00001}[\/latex]<\/th>\n<th>[latex]b[\/latex]<\/th>\n<th>[latex]\\frac{b^{-0.00001}-1}{-0.00001} < B^{\\prime}(0) < \\frac{b^{0.00001}-1}{0.00001}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]0.693145 < B^{\\prime}(0) < 0.69315[\/latex]<\/td>\n<td>[latex]2.7183[\/latex]<\/td>\n<td>[latex]1.000002 < B^{\\prime}(0) < 1.000012[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]2.7[\/latex]<\/td>\n<td>[latex]0.993247 < B^{\\prime}(0) < 0.993257[\/latex]<\/td>\n<td>[latex]2.719[\/latex]<\/td>\n<td>[latex]1.000259 < B^{\\prime}(0) < 1.000269[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]2.71[\/latex]<\/td>\n<td>[latex]0.996944 < B^{\\prime}(0) < 0.996954[\/latex]<\/td>\n<td>[latex]2.72[\/latex]<\/td>\n<td>[latex]1.000627 < B^{\\prime}(0) < 1.000637[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]2.718[\/latex]<\/td>\n<td>[latex]0.999891 < B^{\\prime}(0) < 0.999901[\/latex]<\/td>\n<td>[latex]2.8[\/latex]<\/td>\n<td>[latex]1.029614 < B^{\\prime}(0) < 1.029625[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]2.7182[\/latex]<\/td>\n<td>[latex]0.999965 < B^{\\prime}(0) < 0.999975[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]1.098606 < B^{\\prime}(0) < 1.098618[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1169737978597\">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.<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738187154\">Find the derivative of [latex]f(x)=e^{\\tan (2x)}[\/latex].<\/p>\n<p><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<p><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\">\n<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><br \/>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169737151943\">A colony of mosquitoes has an initial population of [latex]1000.[\/latex] After [latex]t[\/latex] days, the population is given by [latex]A(t)=1000e^{0.3t}[\/latex]. Show that the ratio of the rate of change of the population, [latex]A^{\\prime}(t)[\/latex], to the population size, [latex]A(t)[\/latex] is constant.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738212487\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738212487\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738212487\">First find [latex]A^{\\prime}(t)[\/latex]. By using the chain rule, we have [latex]A^{\\prime}(t)=300e^{0.3t}[\/latex]. Thus, the ratio of the rate of change of the population to the population size is given by<\/p>\n<div id=\"fs-id1169738227689\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\large \\frac{A^{\\prime}(t)}{A(t)} \\normalsize = \\large \\frac{300e^{0.3t}}{1000e^{0.3t}}=0.3[\/latex]<\/div>\n<p id=\"fs-id1169737145031\">The ratio of the rate of change of the population to the population size is the constant [latex]0.3[\/latex].<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=163&amp;end=240&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\/3.9DerivativesOfExponentialAndLogarithmic163to240_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.9 Derivatives of Exponential and Logarithmic Functions&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":23,"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\":\"3.9 Derivatives of Exponential and Logarithmic Functions\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":560,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"3.9 Derivatives of Exponential and Logarithmic Functions","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/264"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":19,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/264\/revisions"}],"predecessor-version":[{"id":4537,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/264\/revisions\/4537"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/560"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/264\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=264"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=264"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=264"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=264"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}