{"id":234,"date":"2023-09-20T22:48:28","date_gmt":"2023-09-20T22:48:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivative-functions\/"},"modified":"2024-08-05T12:35:25","modified_gmt":"2024-08-05T12:35:25","slug":"the-derivative-as-a-function-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-derivative-as-a-function-learn-it-1\/","title":{"raw":"The Derivative as a Function: Learn It 1","rendered":"The Derivative as a Function: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Find the derivative of a function<\/li>\r\n\t<li>Draw the derivative's graph using the original function\u2019s graph<\/li>\r\n\t<li>Explain what it means for a function to be differentiable and how this is connected to being continuous<\/li>\r\n\t<li>Calculate derivatives beyond the first order<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivative Functions<\/h2>\r\n<p>As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious.<\/p>\r\n<p>The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">derivative function<\/h3>\r\n<p id=\"fs-id1169737951873\">Let [latex]f[\/latex] be a function. The <strong>derivative function<\/strong>, denoted by [latex]f^{\\prime}[\/latex], is the function whose domain consists of those values of [latex]x[\/latex] such that the following limit exists:<\/p>\r\n<div id=\"fs-id1169738143059\" class=\"equation\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{f(x+h)-f(x)}{h}[\/latex]<\/div>\r\n<\/section>\r\n<section class=\"textbox proTip\">\r\n<p>A function [latex]f(x)[\/latex] is said to be <strong>differentiable at [latex]a[\/latex]<\/strong> if [latex]f^{\\prime}(a)[\/latex] exists. More generally, a function is said to be <strong>differentiable on [latex]S[\/latex]<\/strong> if it is differentiable at every point in an open set [latex]S[\/latex], and a <strong>differentiable function<\/strong> is one in which [latex]f^{\\prime}(x)[\/latex] exists on its domain.<\/p>\r\n<\/section>\r\n<p id=\"fs-id1169738146367\">In the next few examples we use the definition to find the derivative of a function. There are a few algebraic techniques that are commonly used when using this definition. It may be useful to recall these techniques.<\/p>\r\n<section class=\"textbox recall\">\r\n<ul>\r\n\t<li><strong>Greatest Common Factor (GCF)<\/strong>: The GCF of a polynomial is the largest polynomial that divides evenly into each term of the polynomial. When using the difference quotient, you will often need to factor out a GCF of [latex]h[\/latex]<\/li>\r\n\t<li><strong>Conjugate<\/strong>: For a numerator or a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate, which is found by changing the sign of the radical portion of the denominator. If the denominator is [latex]a+b\\sqrt{c}[\/latex], then the conjugate is [latex]a-b\\sqrt{c}[\/latex].<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169737849750\">Find the derivative of [latex]f(x)=\\sqrt{x}[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169738153091\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169738153091\"]<\/p>\r\n<p id=\"fs-id1169738153091\">Start directly with the definition of the derivative function. Use the definition.<\/p>\r\n<div id=\"fs-id1169737769941\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}f^{\\prime}(x)&amp; =\\underset{h\\to 0}{\\lim}\\frac{\\sqrt{x+h}-\\sqrt{x}}{h} &amp; &amp; &amp; \\begin{array}{l}\\text{Substitute} \\, f(x+h)=\\sqrt{x+h} \\, \\text{and} \\, f(x)=\\sqrt{x} \\\\ \\text{into} \\, f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\frac{f(x+h)-f(x)}{h}. \\end{array} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{\\sqrt{x+h}-\\sqrt{x}}{h}\\cdot \\frac{\\sqrt{x+h}+\\sqrt{x}}{\\sqrt{x+h}+\\sqrt{x}} &amp; &amp; &amp; \\begin{array}{l}\\text{Multiply numerator and denominator by} \\\\ \\sqrt{x+h}+\\sqrt{x} \\, \\text{without distributing in the} \\\\ \\text{denominator.} \\end{array} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{h}{h(\\sqrt{x+h}+\\sqrt{x})} &amp; &amp; &amp; \\text{Multiply the numerators and simplify.} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{1}{(\\sqrt{x+h}+\\sqrt{x})} &amp; &amp; &amp; \\text{Cancel the} \\, h. \\\\ &amp; =\\frac{1}{2\\sqrt{x}} &amp; &amp; &amp; \\text{Evaluate the limit.} \\end{array}[\/latex]<br \/>\r\n[\/hidden-answer]<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169737739857\">Find the derivative of the function [latex]f(x)=x^2-2x[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169737820333\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169737820333\"]<\/p>\r\n<p id=\"fs-id1169737820333\">Follow the same procedure here, but without having to multiply by the conjugate.<\/p>\r\n<div id=\"fs-id1169737946879\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}f^{\\prime}(x) &amp; =\\underset{h\\to 0}{\\lim}\\frac{((x+h)^2-2(x+h))-(x^2-2x)}{h} &amp; &amp; &amp; \\begin{array}{l}\\text{Substitute} \\, f(x+h)=(x+h)^2-2(x+h) \\, \\text{and} \\\\ f(x)=x^2-2x \\, \\text{into} \\\\ f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\frac{f(x+h)-f(x)}{h}. \\end{array} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{x^2+2xh+h^2-2x-2h-x^2+2x}{h} &amp; &amp; &amp; \\text{Expand} \\, (x+h)^2-2(x+h). \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{2xh-2h+h^2}{h} &amp; &amp; &amp; \\text{Simplify.} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{h(2x-2+h)}{h} &amp; &amp; &amp; \\text{Factor out} \\, h \\, \\text{from the numerator.} \\\\ &amp; =\\underset{h\\to 0}{\\lim}(2x-2+h) &amp; &amp; &amp; \\text{Cancel the common factor of} \\, h. \\\\ &amp; =2x-2 &amp; &amp; &amp; \\text{Evaluate the limit.} \\end{array}[\/latex]<br \/>\r\n[\/hidden-answer]<\/div>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288379[\/ohm_question]<\/p>\r\n<\/section>\r\n<section class=\"textbox proTip\">\r\n<p id=\"fs-id1169737950709\">We use a variety of different notations to express the derivative of a function. In the previous example, we showed that if [latex]f(x)=x^2-2x[\/latex], then [latex]f^{\\prime}(x)=2x-2[\/latex].<br \/>\r\n<br \/>\r\nIf we had expressed this function in the form [latex]y=x^2-2x[\/latex], we could have expressed the derivative as [latex]y^{\\prime}=2x-2[\/latex] or [latex]\\frac{dy}{dx}=2x-2[\/latex]. We could have conveyed the same information by writing [latex]\\frac{d}{dx}(x^2-2x)=2x-2[\/latex]. <br \/>\r\n<br \/>\r\nThus, for the function [latex]y=f(x)[\/latex], each of the following notations represents the derivative of [latex]f(x)[\/latex]:<\/p>\r\n<div id=\"fs-id1169738144185\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x), \\,\\, \\dfrac{dy}{dx}, \\,\\, y^{\\prime}, \\,\\, \\dfrac{d}{dx}(f(x))[\/latex]<\/div>\r\n<\/section>\r\n<section class=\"textbox connectIt\">\r\n<p id=\"fs-id1169738219110\">In place of [latex]f^{\\prime}(a)[\/latex] we may also use [latex]\\frac{dy}{dx}\\Big|_{x=a}[\/latex]. Using the [latex]\\frac{dy}{dx}[\/latex] notation (called Leibniz notation) is quite common in engineering and physics. <br \/>\r\n<br \/>\r\nTo understand this notation better, recall that the derivative of a function at a point is the limit of the slopes of secant lines as the secant lines approach the tangent line. <br \/>\r\n<br \/>\r\nThe slopes of these secant lines are often expressed in the form [latex]\\frac{\\Delta y}{\\Delta x}[\/latex] where [latex]\\Delta y[\/latex] is the difference in the [latex]y[\/latex] values corresponding to the difference in the [latex]x[\/latex] values, which are expressed as [latex]\\Delta x[\/latex] (Figure 1). <br \/>\r\n<br \/>\r\nThus the derivative, which can be thought of as the instantaneous rate of change of [latex]y[\/latex] with respect to [latex]x[\/latex], is expressed as<\/p>\r\n<div id=\"fs-id1169738041612\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{dy}{dx}=\\underset{\\Delta x\\to 0}{\\lim}\\dfrac{\\Delta y}{\\Delta x}[\/latex]<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"486\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205220\/CNX_Calc_Figure_03_02_001.jpg\" alt=\"The function y = f(x) is graphed and it shows up as a curve in the first quadrant. The x-axis is marked with 0, a, and a + \u0394x. The y-axis is marked with 0, f(a), and f(a) + \u0394y. There is a straight line crossing y = f(x) at (a, f(a)) and (a + \u0394x, f(a) + \u0394y). From the point (a, f(a)), a horizontal line is drawn; from the point (a + \u0394x, f(a) + \u0394y), a vertical line is drawn. The distance from (a, f(a)) to (a + \u0394x, f(a)) is denoted \u0394x; the distance from (a + \u0394x, f(a) + \u0394y) to (a + \u0394x, f(a)) is denoted \u0394y.\" width=\"486\" height=\"459\" \/> Figure 1. The derivative is expressed as [latex]\\frac{dy}{dx}=\\underset{\\Delta x\\to 0}{\\lim}\\frac{\\Delta y}{\\Delta x}[\/latex].[\/caption]\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Find the derivative of a function<\/li>\n<li>Draw the derivative&#8217;s graph using the original function\u2019s graph<\/li>\n<li>Explain what it means for a function to be differentiable and how this is connected to being continuous<\/li>\n<li>Calculate derivatives beyond the first order<\/li>\n<\/ul>\n<\/section>\n<h2>Derivative Functions<\/h2>\n<p>As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious.<\/p>\n<p>The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">derivative function<\/h3>\n<p id=\"fs-id1169737951873\">Let [latex]f[\/latex] be a function. The <strong>derivative function<\/strong>, denoted by [latex]f^{\\prime}[\/latex], is the function whose domain consists of those values of [latex]x[\/latex] such that the following limit exists:<\/p>\n<div id=\"fs-id1169738143059\" class=\"equation\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{f(x+h)-f(x)}{h}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox proTip\">\n<p>A function [latex]f(x)[\/latex] is said to be <strong>differentiable at [latex]a[\/latex]<\/strong> if [latex]f^{\\prime}(a)[\/latex] exists. More generally, a function is said to be <strong>differentiable on [latex]S[\/latex]<\/strong> if it is differentiable at every point in an open set [latex]S[\/latex], and a <strong>differentiable function<\/strong> is one in which [latex]f^{\\prime}(x)[\/latex] exists on its domain.<\/p>\n<\/section>\n<p id=\"fs-id1169738146367\">In the next few examples we use the definition to find the derivative of a function. There are a few algebraic techniques that are commonly used when using this definition. It may be useful to recall these techniques.<\/p>\n<section class=\"textbox recall\">\n<ul>\n<li><strong>Greatest Common Factor (GCF)<\/strong>: The GCF of a polynomial is the largest polynomial that divides evenly into each term of the polynomial. When using the difference quotient, you will often need to factor out a GCF of [latex]h[\/latex]<\/li>\n<li><strong>Conjugate<\/strong>: For a numerator or a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate, which is found by changing the sign of the radical portion of the denominator. If the denominator is [latex]a+b\\sqrt{c}[\/latex], then the conjugate is [latex]a-b\\sqrt{c}[\/latex].<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169737849750\">Find the derivative of [latex]f(x)=\\sqrt{x}[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738153091\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738153091\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738153091\">Start directly with the definition of the derivative function. Use the definition.<\/p>\n<div id=\"fs-id1169737769941\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}f^{\\prime}(x)& =\\underset{h\\to 0}{\\lim}\\frac{\\sqrt{x+h}-\\sqrt{x}}{h} & & & \\begin{array}{l}\\text{Substitute} \\, f(x+h)=\\sqrt{x+h} \\, \\text{and} \\, f(x)=\\sqrt{x} \\\\ \\text{into} \\, f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\frac{f(x+h)-f(x)}{h}. \\end{array} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{\\sqrt{x+h}-\\sqrt{x}}{h}\\cdot \\frac{\\sqrt{x+h}+\\sqrt{x}}{\\sqrt{x+h}+\\sqrt{x}} & & & \\begin{array}{l}\\text{Multiply numerator and denominator by} \\\\ \\sqrt{x+h}+\\sqrt{x} \\, \\text{without distributing in the} \\\\ \\text{denominator.} \\end{array} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{h}{h(\\sqrt{x+h}+\\sqrt{x})} & & & \\text{Multiply the numerators and simplify.} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{1}{(\\sqrt{x+h}+\\sqrt{x})} & & & \\text{Cancel the} \\, h. \\\\ & =\\frac{1}{2\\sqrt{x}} & & & \\text{Evaluate the limit.} \\end{array}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169737739857\">Find the derivative of the function [latex]f(x)=x^2-2x[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169737820333\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169737820333\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737820333\">Follow the same procedure here, but without having to multiply by the conjugate.<\/p>\n<div id=\"fs-id1169737946879\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll}f^{\\prime}(x) & =\\underset{h\\to 0}{\\lim}\\frac{((x+h)^2-2(x+h))-(x^2-2x)}{h} & & & \\begin{array}{l}\\text{Substitute} \\, f(x+h)=(x+h)^2-2(x+h) \\, \\text{and} \\\\ f(x)=x^2-2x \\, \\text{into} \\\\ f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\frac{f(x+h)-f(x)}{h}. \\end{array} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{x^2+2xh+h^2-2x-2h-x^2+2x}{h} & & & \\text{Expand} \\, (x+h)^2-2(x+h). \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{2xh-2h+h^2}{h} & & & \\text{Simplify.} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{h(2x-2+h)}{h} & & & \\text{Factor out} \\, h \\, \\text{from the numerator.} \\\\ & =\\underset{h\\to 0}{\\lim}(2x-2+h) & & & \\text{Cancel the common factor of} \\, h. \\\\ & =2x-2 & & & \\text{Evaluate the limit.} \\end{array}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288379\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288379&theme=lumen&iframe_resize_id=ohm288379&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<section class=\"textbox proTip\">\n<p id=\"fs-id1169737950709\">We use a variety of different notations to express the derivative of a function. In the previous example, we showed that if [latex]f(x)=x^2-2x[\/latex], then [latex]f^{\\prime}(x)=2x-2[\/latex].<\/p>\n<p>If we had expressed this function in the form [latex]y=x^2-2x[\/latex], we could have expressed the derivative as [latex]y^{\\prime}=2x-2[\/latex] or [latex]\\frac{dy}{dx}=2x-2[\/latex]. We could have conveyed the same information by writing [latex]\\frac{d}{dx}(x^2-2x)=2x-2[\/latex]. <\/p>\n<p>Thus, for the function [latex]y=f(x)[\/latex], each of the following notations represents the derivative of [latex]f(x)[\/latex]:<\/p>\n<div id=\"fs-id1169738144185\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x), \\,\\, \\dfrac{dy}{dx}, \\,\\, y^{\\prime}, \\,\\, \\dfrac{d}{dx}(f(x))[\/latex]<\/div>\n<\/section>\n<section class=\"textbox connectIt\">\n<p id=\"fs-id1169738219110\">In place of [latex]f^{\\prime}(a)[\/latex] we may also use [latex]\\frac{dy}{dx}\\Big|_{x=a}[\/latex]. Using the [latex]\\frac{dy}{dx}[\/latex] notation (called Leibniz notation) is quite common in engineering and physics. <\/p>\n<p>To understand this notation better, recall that the derivative of a function at a point is the limit of the slopes of secant lines as the secant lines approach the tangent line. <\/p>\n<p>The slopes of these secant lines are often expressed in the form [latex]\\frac{\\Delta y}{\\Delta x}[\/latex] where [latex]\\Delta y[\/latex] is the difference in the [latex]y[\/latex] values corresponding to the difference in the [latex]x[\/latex] values, which are expressed as [latex]\\Delta x[\/latex] (Figure 1). <\/p>\n<p>Thus the derivative, which can be thought of as the instantaneous rate of change of [latex]y[\/latex] with respect to [latex]x[\/latex], is expressed as<\/p>\n<div id=\"fs-id1169738041612\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{dy}{dx}=\\underset{\\Delta x\\to 0}{\\lim}\\dfrac{\\Delta y}{\\Delta x}[\/latex]<\/div>\n<figure style=\"width: 486px\" 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\/11205220\/CNX_Calc_Figure_03_02_001.jpg\" alt=\"The function y = f(x) is graphed and it shows up as a curve in the first quadrant. The x-axis is marked with 0, a, and a + \u0394x. The y-axis is marked with 0, f(a), and f(a) + \u0394y. There is a straight line crossing y = f(x) at (a, f(a)) and (a + \u0394x, f(a) + \u0394y). From the point (a, f(a)), a horizontal line is drawn; from the point (a + \u0394x, f(a) + \u0394y), a vertical line is drawn. The distance from (a, f(a)) to (a + \u0394x, f(a)) is denoted \u0394x; the distance from (a + \u0394x, f(a) + \u0394y) to (a + \u0394x, f(a)) is denoted \u0394y.\" width=\"486\" height=\"459\" \/><figcaption class=\"wp-caption-text\">Figure 1. The derivative is expressed as [latex]\\frac{dy}{dx}=\\underset{\\Delta x\\to 0}{\\lim}\\frac{\\Delta y}{\\Delta x}[\/latex].<\/figcaption><\/figure>\n<\/section>\n","protected":false},"author":6,"menu_order":10,"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\":\"3.2 The Derivative as a Function\",\"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":503,"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":"3.2 The Derivative as a Function","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\/234"}],"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":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/234\/revisions"}],"predecessor-version":[{"id":4493,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/234\/revisions\/4493"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/503"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/234\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=234"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=234"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=234"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=234"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}