{"id":267,"date":"2025-02-13T22:45:48","date_gmt":"2025-02-13T22:45:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/derivatives\/"},"modified":"2025-10-24T17:51:35","modified_gmt":"2025-10-24T17:51:35","slug":"derivatives","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/derivatives\/","title":{"raw":"Derivatives: Learn It 1","rendered":"Derivatives: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Find the derivative of a function.<\/li>\r\n \t<li style=\"font-weight: 400;\">Find instantaneous rates of change.<\/li>\r\n \t<li style=\"font-weight: 400;\">Find an equation of the tangent line to the graph of a function at a point.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Finding the Average Rate of Change of a Function<\/h2>\r\nThe functions describing the examples above involve a change over time. Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.\r\n\r\nA <strong>tangent line<\/strong> to a curve is a line that intersects the curve at only a single point but does not cross it there. (The tangent line may intersect the curve at another point away from the point of interest.) If we zoom in on a curve at that point, the curve appears linear, and the <strong>slope of the curve<\/strong> at that point is close to the slope of the tangent line at that point.\r\n\r\nConsider the graph of the function [latex]f\\left(x\\right)={x}^{3}-4x[\/latex]. We can see the slope at various points along the curve.\r\n<div>\r\n<ul>\r\n \t<li>slope at [latex]x=-2[\/latex] is 8<\/li>\r\n \t<li>slope at [latex]x=-1[\/latex] is \u20131<\/li>\r\n \t<li>slope at [latex]x=2[\/latex] is 8<\/li>\r\n<\/ul>\r\n<\/div>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27185355\/CNX_PreCalc_Figure_12_04_0012.jpg\" alt=\"Graph of f(x) = x^3 - 4x with tangent lines at x = -2 with a slope of 8, at x = -3 with a slope of -1, and at x=2 with a slope of 8.\" width=\"487\" height=\"437\" \/>\r\n\r\nGraph showing tangents to curve at \u20132, \u20131, and 2.\r\n\r\nLet\u2019s imagine a point on the curve of function [latex]f[\/latex] at [latex]x=a[\/latex]. The coordinates of the point are [latex]\\left(a,f\\left(a\\right)\\right)[\/latex]. Connect this point with a second point on the curve a little to the right of [latex]x=a[\/latex], with an <em>x<\/em>-value increased by some small real number [latex]h[\/latex]. The coordinates of this second point are [latex]\\left(a+h,f\\left(a+h\\right)\\right)[\/latex] for some positive-value [latex]h[\/latex].\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\/3675\/2018\/09\/27185357\/CNX_PreCalc_Figure_12_04_0022.jpg\" alt=\"Graph of an increasing function that demonstrates the rate of change of the function by drawing a line between the two points, (a, f(a)) and (a, f(a+h)).\" width=\"487\" height=\"290\" \/> Connecting point [latex]a[\/latex] with a point just beyond allows us to measure a slope close to that of a tangent line at [latex]x=a[\/latex].[\/caption]We can calculate the slope of the line connecting the two points [latex]\\left(a,f\\left(a\\right)\\right)[\/latex] and [latex]\\left(a+h,f\\left(a+h\\right)\\right)[\/latex], called a <strong>secant line<\/strong>, by applying the slope formula,\r\n<p style=\"text-align: center;\">[latex]\\text{slope = }\\frac{\\text{change in }y}{\\text{change in }x}[\/latex]<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\text{slope = }\\frac{\\text{change in }y}{\\text{change in }x}[\/latex]<\/div>\r\nWe use the notation [latex]{m}_{\\sec }[\/latex] to represent the slope of the secant line connecting two points.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{m}_{\\sec }&amp;=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{\\left(a+h\\right)-\\left(a\\right)} \\\\ &amp;=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h} \\end{align}[\/latex]<\/div>\r\nThe slope [latex]{m}_{\\sec }[\/latex] equals the <em>average rate of change<\/em> between two points [latex]\\left(a,f\\left(a\\right)\\right)[\/latex] and [latex]\\left(a+h,f\\left(a+h\\right)\\right)[\/latex].\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>average rate of change<\/h3>\r\nThe average rate of change (AROC) between two points [latex]\\left(a,f\\left(a\\right)\\right)[\/latex] and [latex]\\left(a+h,f\\left(a+h\\right)\\right)[\/latex] on the curve of [latex]f[\/latex] is the slope of the line connecting the two points and is given by\r\n<p style=\"text-align: center;\">[latex]\\text{AROC}=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the average rate of change connecting the points [latex]\\left(2,-6\\right)[\/latex] and [latex]\\left(-1,5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"928775\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"928775\"]\r\n\r\nWe know the average rate of change connecting two points may be given by\r\n<p style=\"text-align: center;\">[latex]\\text{AROC}=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex].<\/p>\r\nIf one point is [latex]\\left(2,-6\\right)[\/latex], or [latex]\\left(2,f\\left(2\\right)\\right)[\/latex], then [latex]f\\left(2\\right)=-6[\/latex].\r\n\r\nThe value [latex]h[\/latex] is the displacement from [latex]2[\/latex] to [latex]-1[\/latex], which equals [latex]-1 - 2=-3[\/latex].\r\n\r\nFor the other point, [latex]f\\left(a+h\\right)[\/latex] is the <em>y<\/em>-coordinate at [latex]a+h[\/latex], which is [latex]2+\\left(-3\\right)[\/latex] or [latex]-1[\/latex], so [latex]f\\left(a+h\\right)=f\\left(-1\\right)=5[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\text{AROC}&amp;=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h} \\\\ &amp;=\\frac{5-\\left(-6\\right)}{-3} \\\\ &amp;=\\frac{11}{-3} \\\\ &amp;=-\\frac{11}{3} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Find the average rate of change connecting the points [latex]\\left(-5,1.5\\right)[\/latex] and [latex]\\left(-2.5,9\\right)[\/latex].[reveal-answer q=\"459741\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"459741\"]3[\/hidden-answer]<\/section>\r\n<h2>Understanding the Instantaneous Rate of Change<\/h2>\r\nWhen calculating average rate of change we find the change between two points on the curve of a function. Suppose we make the distance between the two points closer together. As the second point approaches the first point, the connecting line between the two points, called the secant line, will get closer and closer to being a tangent to the function at [latex]x=a[\/latex], and the slope of the secant line will get closer and closer to the slope of the tangent at [latex]x=a[\/latex].\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27185359\/CNX_Precalc_Figure_12_04_0032.jpg\" alt=\"Graph of an increasing function that contains a point, P, at (a, f(a)). At the point, there is a tangent line and two secant lines where one secant line is connected to Q1 and another secant line is connected to Q2.\" width=\"487\" height=\"401\" \/>\r\n\r\nThe connecting line between two points moves closer to being a tangent line at [latex]x=a[\/latex].\r\n\r\nBecause we are looking for the <strong>slope of the tangent<\/strong> at [latex]x=a[\/latex], we can think of the measure of the slope of the curve of a function [latex]f[\/latex] at a given point as the rate of change at a particular instant. We call this slope the <strong>instantaneous rate of change<\/strong>, or the <strong>derivative<\/strong> of the function at [latex]x=a[\/latex]. Both can be found by finding the limit of the slope of a line connecting the point at [latex]x=a[\/latex] with a second point infinitesimally close along the curve. For a function [latex]f[\/latex] both the instantaneous rate of change of the function and the derivative of the function at [latex]x=a[\/latex] are written as [latex]f\\text{'}\\left(a\\right)[\/latex], and we can define them as a <strong>two-sided limit<\/strong> that has the same value whether approached from the left or the right.\r\n<div style=\"text-align: center;\">[latex]{f}^{\\prime }\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/div>\r\nThe expression by which the limit is found is known as the <strong>difference quotient<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>derivative<\/h3>\r\nThe <strong>derivative<\/strong>, or <strong>instantaneous rate of change<\/strong>, of a function [latex]f[\/latex] at [latex]x=a[\/latex], is given by\r\n<p style=\"text-align: center;\">[latex]{f}^{\\prime }\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/p>\r\nThe expression [latex]\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex] is called the difference quotient.\r\n\r\nWe use the difference quotient to evaluate the limit of the rate of change of the function as [latex]h[\/latex] approaches 0.\r\n\r\n<\/section>The <strong>derivative<\/strong> of a function can be interpreted in different ways. It can be observed as the behavior of a graph of the function or calculated as a numerical rate of change of the function.\r\n<ul>\r\n \t<li>The derivative of a function [latex]f\\left(x\\right)[\/latex] at a point [latex]x=a[\/latex] is the slope of the tangent line to the curve [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex]. The derivative of [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex] is written [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex].<\/li>\r\n \t<li>The derivative [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex] measures how the curve changes at the point [latex]\\left(a,f\\left(a\\right)\\right)[\/latex].<\/li>\r\n \t<li>The derivative [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex] may be thought of as the instantaneous rate of change of the function [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex].<\/li>\r\n \t<li>If a function measures distance as a function of time, then the derivative measures the instantaneous <strong>velocity<\/strong> at time [latex]t=a[\/latex].<\/li>\r\n<\/ul>\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">The equation of the derivative of a function [latex]f\\left(x\\right)[\/latex] is written as [latex]\\begin{align}{y}^{\\prime }={f}^{\\prime }\\left(x\\right)\\end{align}[\/latex], where [latex]y=f\\left(x\\right)[\/latex]. The notation [latex]\\begin{align}{f}^{\\prime }\\left(x\\right)\\end{align}[\/latex] is read as \" [latex]f\\text{ prime of }x[\/latex]. \" Alternate notations for the derivative include the following:[latex]\\begin{align}{f}^{\\prime }\\left(x\\right)={y}^{\\prime }=\\frac{dy}{dx}=\\frac{df}{dx}=\\frac{d}{dx}f\\left(x\\right)=Df\\left(x\\right)\\end{align}[\/latex]The expression [latex]\\begin{align}{f}^{\\prime }\\left(x\\right)\\end{align}[\/latex] is now a function of [latex]x[\/latex] ; this function gives the slope of the curve [latex]y=f\\left(x\\right)[\/latex] at any value of [latex]x[\/latex]. The derivative of a function [latex]f\\left(x\\right)[\/latex] at a point [latex]x=a[\/latex] is denoted [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex].<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a function [latex]f[\/latex], find the derivative by applying the definition of the derivative.<\/strong>\r\n<ol>\r\n \t<li>Calculate [latex]f\\left(a+h\\right)[\/latex].<\/li>\r\n \t<li>Calculate [latex]f\\left(a\\right)[\/latex].<\/li>\r\n \t<li>Substitute and simplify [latex]\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex].<\/li>\r\n \t<li>Evaluate the limit if it exists: [latex]{f}^{\\prime }\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the derivative of the function [latex]f\\left(x\\right)={x}^{2}-3x+5[\/latex] at [latex]x=a[\/latex].[reveal-answer q=\"712454\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"712454\"]We have:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{f}^{\\prime }\\left(a\\right)&amp;=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h} &amp;&amp; \\text{ Definition of a derivative } \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]f\\left(a+h\\right)={\\left(a+h\\right)}^{2}-3\\left(a+h\\right)+5[\/latex] and [latex]f\\left(a\\right)={a}^{2}-3a+5[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{f}^{\\prime }\\left(a\\right)&amp;=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\left(a+h\\right)\\left(a+h\\right)-3\\left(a+h\\right)+5-\\left({a}^{2}-3a+5\\right)}{h} \\\\ &amp;=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{{a}^{2}+2ah+{h}^{2}-3a - 3h+5-{a}^{2}+3a - 5}{h} &amp;&amp; \\text{Evaluate to remove parentheses}. \\\\ &amp;=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\cancel{{a}^{2}}+2ah+{h}^{2}\\cancel{-3a}-3h\\cancel{+5}\\cancel{-{a}^{2}}\\cancel{+3a}\\cancel{-5}}{h}&amp;&amp; \\text{Simplify}. \\\\ &amp;=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{2ah+{h}^{2}-3h}{h} \\\\ &amp;=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\cancel{h}\\left(2a+h - 3\\right)}{\\cancel{h}}&amp;&amp; \\text{Factor out an }h. \\\\ &amp;=2a+0 - 3&amp;&amp; \\text{Evaluate the limit}. \\\\ &amp;=2a - 3 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Find the derivative of the function [latex]f\\left(x\\right)=3{x}^{2}+7x[\/latex] at [latex]x=a[\/latex].[reveal-answer q=\"274812\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"274812\"][latex]\\begin{align}{f}^{\\prime }\\left(a\\right)=6a+7\\end{align}[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]16117[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li style=\"font-weight: 400;\">Find the derivative of a function.<\/li>\n<li style=\"font-weight: 400;\">Find instantaneous rates of change.<\/li>\n<li style=\"font-weight: 400;\">Find an equation of the tangent line to the graph of a function at a point.<\/li>\n<\/ul>\n<\/section>\n<h2>Finding the Average Rate of Change of a Function<\/h2>\n<p>The functions describing the examples above involve a change over time. Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.<\/p>\n<p>A <strong>tangent line<\/strong> to a curve is a line that intersects the curve at only a single point but does not cross it there. (The tangent line may intersect the curve at another point away from the point of interest.) If we zoom in on a curve at that point, the curve appears linear, and the <strong>slope of the curve<\/strong> at that point is close to the slope of the tangent line at that point.<\/p>\n<p>Consider the graph of the function [latex]f\\left(x\\right)={x}^{3}-4x[\/latex]. We can see the slope at various points along the curve.<\/p>\n<div>\n<ul>\n<li>slope at [latex]x=-2[\/latex] is 8<\/li>\n<li>slope at [latex]x=-1[\/latex] is \u20131<\/li>\n<li>slope at [latex]x=2[\/latex] is 8<\/li>\n<\/ul>\n<\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27185355\/CNX_PreCalc_Figure_12_04_0012.jpg\" alt=\"Graph of f(x) = x^3 - 4x with tangent lines at x = -2 with a slope of 8, at x = -3 with a slope of -1, and at x=2 with a slope of 8.\" width=\"487\" height=\"437\" \/><\/p>\n<p>Graph showing tangents to curve at \u20132, \u20131, and 2.<\/p>\n<p>Let\u2019s imagine a point on the curve of function [latex]f[\/latex] at [latex]x=a[\/latex]. The coordinates of the point are [latex]\\left(a,f\\left(a\\right)\\right)[\/latex]. Connect this point with a second point on the curve a little to the right of [latex]x=a[\/latex], with an <em>x<\/em>-value increased by some small real number [latex]h[\/latex]. The coordinates of this second point are [latex]\\left(a+h,f\\left(a+h\\right)\\right)[\/latex] for some positive-value [latex]h[\/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\/3675\/2018\/09\/27185357\/CNX_PreCalc_Figure_12_04_0022.jpg\" alt=\"Graph of an increasing function that demonstrates the rate of change of the function by drawing a line between the two points, (a, f(a)) and (a, f(a+h)).\" width=\"487\" height=\"290\" \/><figcaption class=\"wp-caption-text\">Connecting point [latex]a[\/latex] with a point just beyond allows us to measure a slope close to that of a tangent line at [latex]x=a[\/latex].<\/figcaption><\/figure>\n<p>We can calculate the slope of the line connecting the two points [latex]\\left(a,f\\left(a\\right)\\right)[\/latex] and [latex]\\left(a+h,f\\left(a+h\\right)\\right)[\/latex], called a <strong>secant line<\/strong>, by applying the slope formula,<\/p>\n<p style=\"text-align: center;\">[latex]\\text{slope = }\\frac{\\text{change in }y}{\\text{change in }x}[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\text{slope = }\\frac{\\text{change in }y}{\\text{change in }x}[\/latex]<\/div>\n<p>We use the notation [latex]{m}_{\\sec }[\/latex] to represent the slope of the secant line connecting two points.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{m}_{\\sec }&=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{\\left(a+h\\right)-\\left(a\\right)} \\\\ &=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h} \\end{align}[\/latex]<\/div>\n<p>The slope [latex]{m}_{\\sec }[\/latex] equals the <em>average rate of change<\/em> between two points [latex]\\left(a,f\\left(a\\right)\\right)[\/latex] and [latex]\\left(a+h,f\\left(a+h\\right)\\right)[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>average rate of change<\/h3>\n<p>The average rate of change (AROC) between two points [latex]\\left(a,f\\left(a\\right)\\right)[\/latex] and [latex]\\left(a+h,f\\left(a+h\\right)\\right)[\/latex] on the curve of [latex]f[\/latex] is the slope of the line connecting the two points and is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\text{AROC}=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the average rate of change connecting the points [latex]\\left(2,-6\\right)[\/latex] and [latex]\\left(-1,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q928775\">Show Solution<\/button><\/p>\n<div id=\"q928775\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know the average rate of change connecting two points may be given by<\/p>\n<p style=\"text-align: center;\">[latex]\\text{AROC}=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex].<\/p>\n<p>If one point is [latex]\\left(2,-6\\right)[\/latex], or [latex]\\left(2,f\\left(2\\right)\\right)[\/latex], then [latex]f\\left(2\\right)=-6[\/latex].<\/p>\n<p>The value [latex]h[\/latex] is the displacement from [latex]2[\/latex] to [latex]-1[\/latex], which equals [latex]-1 - 2=-3[\/latex].<\/p>\n<p>For the other point, [latex]f\\left(a+h\\right)[\/latex] is the <em>y<\/em>-coordinate at [latex]a+h[\/latex], which is [latex]2+\\left(-3\\right)[\/latex] or [latex]-1[\/latex], so [latex]f\\left(a+h\\right)=f\\left(-1\\right)=5[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\text{AROC}&=\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h} \\\\ &=\\frac{5-\\left(-6\\right)}{-3} \\\\ &=\\frac{11}{-3} \\\\ &=-\\frac{11}{3} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Find the average rate of change connecting the points [latex]\\left(-5,1.5\\right)[\/latex] and [latex]\\left(-2.5,9\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q459741\">Show Solution<\/button><\/p>\n<div id=\"q459741\" class=\"hidden-answer\" style=\"display: none\">3<\/div>\n<\/div>\n<\/section>\n<h2>Understanding the Instantaneous Rate of Change<\/h2>\n<p>When calculating average rate of change we find the change between two points on the curve of a function. Suppose we make the distance between the two points closer together. As the second point approaches the first point, the connecting line between the two points, called the secant line, will get closer and closer to being a tangent to the function at [latex]x=a[\/latex], and the slope of the secant line will get closer and closer to the slope of the tangent at [latex]x=a[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27185359\/CNX_Precalc_Figure_12_04_0032.jpg\" alt=\"Graph of an increasing function that contains a point, P, at (a, f(a)). At the point, there is a tangent line and two secant lines where one secant line is connected to Q1 and another secant line is connected to Q2.\" width=\"487\" height=\"401\" \/><\/p>\n<p>The connecting line between two points moves closer to being a tangent line at [latex]x=a[\/latex].<\/p>\n<p>Because we are looking for the <strong>slope of the tangent<\/strong> at [latex]x=a[\/latex], we can think of the measure of the slope of the curve of a function [latex]f[\/latex] at a given point as the rate of change at a particular instant. We call this slope the <strong>instantaneous rate of change<\/strong>, or the <strong>derivative<\/strong> of the function at [latex]x=a[\/latex]. Both can be found by finding the limit of the slope of a line connecting the point at [latex]x=a[\/latex] with a second point infinitesimally close along the curve. For a function [latex]f[\/latex] both the instantaneous rate of change of the function and the derivative of the function at [latex]x=a[\/latex] are written as [latex]f\\text{'}\\left(a\\right)[\/latex], and we can define them as a <strong>two-sided limit<\/strong> that has the same value whether approached from the left or the right.<\/p>\n<div style=\"text-align: center;\">[latex]{f}^{\\prime }\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/div>\n<p>The expression by which the limit is found is known as the <strong>difference quotient<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>derivative<\/h3>\n<p>The <strong>derivative<\/strong>, or <strong>instantaneous rate of change<\/strong>, of a function [latex]f[\/latex] at [latex]x=a[\/latex], is given by<\/p>\n<p style=\"text-align: center;\">[latex]{f}^{\\prime }\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/p>\n<p>The expression [latex]\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex] is called the difference quotient.<\/p>\n<p>We use the difference quotient to evaluate the limit of the rate of change of the function as [latex]h[\/latex] approaches 0.<\/p>\n<\/section>\n<p>The <strong>derivative<\/strong> of a function can be interpreted in different ways. It can be observed as the behavior of a graph of the function or calculated as a numerical rate of change of the function.<\/p>\n<ul>\n<li>The derivative of a function [latex]f\\left(x\\right)[\/latex] at a point [latex]x=a[\/latex] is the slope of the tangent line to the curve [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex]. The derivative of [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex] is written [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex].<\/li>\n<li>The derivative [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex] measures how the curve changes at the point [latex]\\left(a,f\\left(a\\right)\\right)[\/latex].<\/li>\n<li>The derivative [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex] may be thought of as the instantaneous rate of change of the function [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex].<\/li>\n<li>If a function measures distance as a function of time, then the derivative measures the instantaneous <strong>velocity<\/strong> at time [latex]t=a[\/latex].<\/li>\n<\/ul>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">The equation of the derivative of a function [latex]f\\left(x\\right)[\/latex] is written as [latex]\\begin{align}{y}^{\\prime }={f}^{\\prime }\\left(x\\right)\\end{align}[\/latex], where [latex]y=f\\left(x\\right)[\/latex]. The notation [latex]\\begin{align}{f}^{\\prime }\\left(x\\right)\\end{align}[\/latex] is read as &#8221; [latex]f\\text{ prime of }x[\/latex]. &#8221; Alternate notations for the derivative include the following:[latex]\\begin{align}{f}^{\\prime }\\left(x\\right)={y}^{\\prime }=\\frac{dy}{dx}=\\frac{df}{dx}=\\frac{d}{dx}f\\left(x\\right)=Df\\left(x\\right)\\end{align}[\/latex]The expression [latex]\\begin{align}{f}^{\\prime }\\left(x\\right)\\end{align}[\/latex] is now a function of [latex]x[\/latex] ; this function gives the slope of the curve [latex]y=f\\left(x\\right)[\/latex] at any value of [latex]x[\/latex]. The derivative of a function [latex]f\\left(x\\right)[\/latex] at a point [latex]x=a[\/latex] is denoted [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex].<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a function [latex]f[\/latex], find the derivative by applying the definition of the derivative.<\/strong><\/p>\n<ol>\n<li>Calculate [latex]f\\left(a+h\\right)[\/latex].<\/li>\n<li>Calculate [latex]f\\left(a\\right)[\/latex].<\/li>\n<li>Substitute and simplify [latex]\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex].<\/li>\n<li>Evaluate the limit if it exists: [latex]{f}^{\\prime }\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the derivative of the function [latex]f\\left(x\\right)={x}^{2}-3x+5[\/latex] at [latex]x=a[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q712454\">Show Solution<\/button><\/p>\n<div id=\"q712454\" class=\"hidden-answer\" style=\"display: none\">We have:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{f}^{\\prime }\\left(a\\right)&=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h} && \\text{ Definition of a derivative } \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]f\\left(a+h\\right)={\\left(a+h\\right)}^{2}-3\\left(a+h\\right)+5[\/latex] and [latex]f\\left(a\\right)={a}^{2}-3a+5[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{f}^{\\prime }\\left(a\\right)&=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\left(a+h\\right)\\left(a+h\\right)-3\\left(a+h\\right)+5-\\left({a}^{2}-3a+5\\right)}{h} \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{{a}^{2}+2ah+{h}^{2}-3a - 3h+5-{a}^{2}+3a - 5}{h} && \\text{Evaluate to remove parentheses}. \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\cancel{{a}^{2}}+2ah+{h}^{2}\\cancel{-3a}-3h\\cancel{+5}\\cancel{-{a}^{2}}\\cancel{+3a}\\cancel{-5}}{h}&& \\text{Simplify}. \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{2ah+{h}^{2}-3h}{h} \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\cancel{h}\\left(2a+h - 3\\right)}{\\cancel{h}}&& \\text{Factor out an }h. \\\\ &=2a+0 - 3&& \\text{Evaluate the limit}. \\\\ &=2a - 3 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Find the derivative of the function [latex]f\\left(x\\right)=3{x}^{2}+7x[\/latex] at [latex]x=a[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q274812\">Show Solution<\/button><\/p>\n<div id=\"q274812\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align}{f}^{\\prime }\\left(a\\right)=6a+7\\end{align}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm16117\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16117&theme=lumen&iframe_resize_id=ohm16117&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":6,"menu_order":24,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"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":263,"module-header":"learn_it","content_attributions":[{"type":"cc-attribution","description":"Precalculus","author":"OpenStax College","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface","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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/267"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/267\/revisions"}],"predecessor-version":[{"id":4898,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/267\/revisions\/4898"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/263"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/267\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=267"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=267"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=267"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=267"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}