{"id":331,"date":"2021-02-04T01:12:58","date_gmt":"2021-02-04T01:12:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=331"},"modified":"2022-03-16T05:27:55","modified_gmt":"2022-03-16T05:27:55","slug":"derivative-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/derivative-functions\/","title":{"raw":"Derivative Functions","rendered":"Derivative Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Define the derivative function of a given function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1169738213881\" class=\"bc-section section\">\r\n<p id=\"fs-id1169737807119\">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\r\n<div id=\"fs-id1169737947145\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\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\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&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1169738193023\">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<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\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Factoring out a greatest common factor<\/h3>\r\nThe <strong>greatest common factor<\/strong> (GCF) of a polynomial is the largest polynomial that divides evenly into each term of the polynomial.\r\n\r\nWhen using the difference quotient, there will be many times where you will need to factor out a GCF of\u00a0<em>h.\u00a0<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Rationalizing a numerator or denominator with rational and irrational terms<\/h3>\r\nFor 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].\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738213881\" class=\"bc-section section\">\r\n<div id=\"fs-id1169737770972\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Derivative of a Square-Root Function<\/h3>\r\n<p id=\"fs-id1169737849750\">Find the derivative of [latex]f(x)=\\sqrt{x}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169738153091\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738153091\"]\r\n<p id=\"fs-id1169738153091\">Start directly with the definition of the derivative function. Use the definition.<\/p>\r\n\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]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737774001\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Derivative of a Quadratic Function<\/h3>\r\n<p id=\"fs-id1169737739857\">Find the derivative of the function [latex]f(x)=x^2-2x[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169737820333\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737820333\"]\r\n<p id=\"fs-id1169737820333\">Follow the same procedure here, but without having to multiply by the conjugate.<\/p>\r\n\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]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738007264\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169737819520\">Find the derivative of [latex]f(x)=x^2[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169737789025\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737789025\"]\r\n<p id=\"fs-id1169737789025\">[latex]f^{\\prime}(x)=2x[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169737812479\">Use the definition and follow the example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/w9m4h4i47-M?controls=0&amp;start=349&amp;end=408&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.2TheDerivativeAsAFunction349to408_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.2 The Derivative as a Function\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]16116[\/ohm_question]\r\n\r\n<\/div>\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]. 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]. Thus, for the function [latex]y=f(x)[\/latex], each of the following notations represents the derivative of [latex]f(x)[\/latex]:<\/p>\r\n\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&nbsp;\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] Use of the [latex]\\frac{dy}{dx}[\/latex] notation (called Leibniz notation) is quite common in engineering and physics. 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. 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). 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>\r\n\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&nbsp;\r\n<div><\/div>\r\n<div>[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]<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Define the derivative function of a given function<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1169738213881\" class=\"bc-section section\">\n<p id=\"fs-id1169737807119\">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<div id=\"fs-id1169737947145\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1169738193023\">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<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<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Factoring out a greatest common factor<\/h3>\n<p>The <strong>greatest common factor<\/strong> (GCF) of a polynomial is the largest polynomial that divides evenly into each term of the polynomial.<\/p>\n<p>When using the difference quotient, there will be many times where you will need to factor out a GCF of\u00a0<em>h.\u00a0<\/em><\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Rationalizing a numerator or denominator with rational and irrational terms<\/h3>\n<p>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].<\/p>\n<\/div>\n<div id=\"fs-id1169738213881\" class=\"bc-section section\">\n<div id=\"fs-id1169737770972\" class=\"textbook exercises\">\n<h3>Example: Finding the Derivative of a Square-Root Function<\/h3>\n<p id=\"fs-id1169737849750\">Find the derivative of [latex]f(x)=\\sqrt{x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738153091\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<div id=\"fs-id1169737774001\" class=\"textbook exercises\">\n<h3>Example: Finding the Derivative of a Quadratic Function<\/h3>\n<p id=\"fs-id1169737739857\">Find the derivative of the function [latex]f(x)=x^2-2x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737820333\">Show Solution<\/span><\/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<\/div>\n<div id=\"fs-id1169738007264\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169737819520\">Find the derivative of [latex]f(x)=x^2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737789025\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737789025\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737789025\">[latex]f^{\\prime}(x)=2x[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1169737812479\">Use the definition and follow the example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/w9m4h4i47-M?controls=0&amp;start=349&amp;end=408&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.2TheDerivativeAsAFunction349to408_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.2 The Derivative as a Function&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm16116\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16116&theme=oea&iframe_resize_id=ohm16116&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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]. 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]. 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<p>&nbsp;<\/p>\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] Use of the [latex]\\frac{dy}{dx}[\/latex] notation (called Leibniz notation) is quite common in engineering and physics. 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. 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). 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<p>&nbsp;<\/p>\n<div><\/div>\n<div>\n<div style=\"width: 496px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-331\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.2 The Derivative as a Function. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":7,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-331","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/331","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":22,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/331\/revisions"}],"predecessor-version":[{"id":4803,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/331\/revisions\/4803"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/35"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/331\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=331"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=331"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=331"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=331"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}