{"id":334,"date":"2021-02-04T01:13:34","date_gmt":"2021-02-04T01:13:34","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=334"},"modified":"2024-07-03T15:56:35","modified_gmt":"2024-07-03T15:56:35","slug":"higher-order-derivatives","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/higher-order-derivatives\/","title":{"raw":"Higher-Order Derivatives","rendered":"Higher-Order Derivatives"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Explain the meaning of a higher-order derivative<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169738225641\">The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as <strong>higher-order derivatives<\/strong>. The notation for the higher-order derivatives of [latex]y=f(x)[\/latex] can be expressed in any of the following forms:<\/p>\r\n\r\n<div id=\"fs-id1169738214590\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f''(x), \\, f'''(x), \\, f^{(4)}(x), \\cdots ,f^{(n)}(x)[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div id=\"fs-id1169738221934\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y'', \\, y''', \\, y^{(4)}, \\cdots ,y^{(n)}[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div id=\"fs-id1169738218117\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{d^2y}{dx^2}, \\, \\dfrac{d^3y}{dx^3}, \\, \\dfrac{d^4y}{dx^4}, \\cdots,\\dfrac{d^ny}{dx^n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738214546\" style=\"text-align: left;\">It is interesting to note that the notation for [latex]\\frac{d^2y}{dx^2}[\/latex] may be viewed as an attempt to express [latex]\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)[\/latex] more compactly. Analogously,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left(\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)\\right)=\\frac{d}{dx}\\left(\\frac{d^2y}{dx^2}\\right)=\\frac{d^3y}{dx^3}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1169738217346\" class=\"textbook exercises\">\r\n<h3>Example: Finding a Second Derivative<\/h3>\r\n<p id=\"fs-id1169738217355\">For [latex]f(x)=2x^2-3x+1[\/latex], find [latex]f''(x)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169738219164\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738219164\"]\r\n<p id=\"fs-id1169738219164\">First find [latex]f^{\\prime}(x)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169738219187\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}f^{\\prime}(x) &amp; =\\underset{h\\to 0}{\\lim}\\frac{(2(x+h)^2-3(x+h)+1)-(2x^2-3x+1)}{h} &amp; &amp; &amp; \\begin{array}{l}\\text{Substitute} \\, f(x)=2x^2-3x+1 \\\\ \\text{and} \\\\ f(x+h)=2(x+h)^2-3(x+h)+1 \\\\ \\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{4xh+h^2-3h}{h} &amp; &amp; &amp; \\text{Simplify the numerator.} \\\\ &amp; =\\underset{h\\to 0}{\\lim}(4x+h-3) &amp; &amp; &amp; \\begin{array}{l}\\text{Factor out the} \\, h \\, \\text{in the numerator} \\\\ \\text{and cancel with the} \\, h \\, \\text{in the} \\\\ \\text{denominator.} \\end{array} \\\\ &amp; =4x-3 &amp; &amp; &amp; \\text{Take the limit.} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737142255\">Next, find [latex]f''(x)[\/latex] by taking the derivative of [latex]f^{\\prime}(x)=4x-3[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169738068328\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} f''(x)&amp; =\\underset{h\\to 0}{\\lim}\\frac{f^{\\prime}(x+h)-f^{\\prime}(x)}{h} &amp; &amp; &amp; \\begin{array}{l}\\text{Use} \\, f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\frac{f(x+h)-f(x)}{h} \\, \\text{with} \\, f^{\\prime}(x) \\, \\text{in} \\\\ \\text{place of} \\, f(x). \\end{array} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{(4(x+h)-3)-(4x-3)}{h} &amp; &amp; &amp; \\begin{array}{l}\\text{Substitute} \\, f^{\\prime}(x+h)=4(x+h)-3 \\, \\text{and} \\\\ f^{\\prime}(x)=4x-3. \\end{array} \\\\ &amp; =\\underset{h\\to 0}{\\lim}4 &amp; &amp; &amp; \\text{Simplify.} \\\\ &amp; =4 &amp; &amp; &amp; \\text{Take the limit.} \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738187249\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169738187258\">Find [latex]f''(x)[\/latex] for [latex]f(x)=x^2[\/latex].<\/p>\r\n[reveal-answer q=\"1238970\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"1238970\"]\r\n<p id=\"fs-id1169738099392\">We found [latex]f^{\\prime}(x)=2x[\/latex] in a previous checkpoint. Use the definition to find the derivative of [latex]f^{\\prime}(x)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169738099362\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738099362\"]\r\n<p id=\"fs-id1169738099362\">[latex]f''(x)=2[\/latex]<\/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=1432&amp;end=1510&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\/3.2TheDerivativeAsAFunction1432to1510_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 id=\"fs-id1169738099443\" class=\"textbook exercises\">\r\n<h3>Example: Finding Acceleration<\/h3>\r\n<p id=\"fs-id1169738099452\">The position of a particle along a coordinate axis at time [latex]t[\/latex] (in seconds) is given by [latex]s(t)=3t^2-4t+1[\/latex] (in meters). Find the function that describes its acceleration at time [latex]t[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169738219009\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738219009\"]\r\n<p id=\"fs-id1169738219009\">Since [latex]v(t)=s^{\\prime}(t)[\/latex] and [latex]a(t)=v^{\\prime}(t)=s''(t)[\/latex], we begin by finding the derivative of [latex]s(t)[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1169737911373\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}s^{\\prime}(t) &amp; =\\underset{h\\to 0}{\\lim}\\frac{s(t+h)-s(t)}{h} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{3(t+h)^2-4(t+h)+1-(3t^2-4t+1)}{h} \\\\ &amp; =6t-4 \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737950766\">Next,<\/p>\r\n\r\n<div id=\"fs-id1169737950769\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} s''(t) &amp; =\\underset{h\\to 0}{\\lim}\\frac{s^{\\prime}(t+h)-s^{\\prime}(t)}{h} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{6(t+h)-4-(6t-4)}{h} \\\\ &amp; =6 \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738185118\">Thus, [latex]a=6 \\, \\text{m\/s}^2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Finding Acceleration.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/w9m4h4i47-M?controls=0&amp;start=1513&amp;end=1713&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]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\/3.2TheDerivativeAsAFunction1513to1713_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 id=\"fs-id1169738185145\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169738185153\">For [latex]s(t)=t^3[\/latex], find [latex]a(t)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169737935177\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737935177\"]\r\n<p id=\"fs-id1169737935177\">[latex]a(t)=6t[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169737935205\">Use the previous examples as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Explain the meaning of a higher-order derivative<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169738225641\">The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as <strong>higher-order derivatives<\/strong>. The notation for the higher-order derivatives of [latex]y=f(x)[\/latex] can be expressed in any of the following forms:<\/p>\n<div id=\"fs-id1169738214590\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f''(x), \\, f'''(x), \\, f^{(4)}(x), \\cdots ,f^{(n)}(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div id=\"fs-id1169738221934\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y'', \\, y''', \\, y^{(4)}, \\cdots ,y^{(n)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div id=\"fs-id1169738218117\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{d^2y}{dx^2}, \\, \\dfrac{d^3y}{dx^3}, \\, \\dfrac{d^4y}{dx^4}, \\cdots,\\dfrac{d^ny}{dx^n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738214546\" style=\"text-align: left;\">It is interesting to note that the notation for [latex]\\frac{d^2y}{dx^2}[\/latex] may be viewed as an attempt to express [latex]\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)[\/latex] more compactly. Analogously,<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}\\left(\\frac{d}{dx}\\left(\\frac{dy}{dx}\\right)\\right)=\\frac{d}{dx}\\left(\\frac{d^2y}{dx^2}\\right)=\\frac{d^3y}{dx^3}[\/latex]<\/p>\n<div id=\"fs-id1169738217346\" class=\"textbook exercises\">\n<h3>Example: Finding a Second Derivative<\/h3>\n<p id=\"fs-id1169738217355\">For [latex]f(x)=2x^2-3x+1[\/latex], find [latex]f''(x)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738219164\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738219164\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738219164\">First find [latex]f^{\\prime}(x)[\/latex].<\/p>\n<div id=\"fs-id1169738219187\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}f^{\\prime}(x) & =\\underset{h\\to 0}{\\lim}\\frac{(2(x+h)^2-3(x+h)+1)-(2x^2-3x+1)}{h} & & & \\begin{array}{l}\\text{Substitute} \\, f(x)=2x^2-3x+1 \\\\ \\text{and} \\\\ f(x+h)=2(x+h)^2-3(x+h)+1 \\\\ \\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{4xh+h^2-3h}{h} & & & \\text{Simplify the numerator.} \\\\ & =\\underset{h\\to 0}{\\lim}(4x+h-3) & & & \\begin{array}{l}\\text{Factor out the} \\, h \\, \\text{in the numerator} \\\\ \\text{and cancel with the} \\, h \\, \\text{in the} \\\\ \\text{denominator.} \\end{array} \\\\ & =4x-3 & & & \\text{Take the limit.} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737142255\">Next, find [latex]f''(x)[\/latex] by taking the derivative of [latex]f^{\\prime}(x)=4x-3[\/latex].<\/p>\n<div id=\"fs-id1169738068328\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} f''(x)& =\\underset{h\\to 0}{\\lim}\\frac{f^{\\prime}(x+h)-f^{\\prime}(x)}{h} & & & \\begin{array}{l}\\text{Use} \\, f^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\frac{f(x+h)-f(x)}{h} \\, \\text{with} \\, f^{\\prime}(x) \\, \\text{in} \\\\ \\text{place of} \\, f(x). \\end{array} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{(4(x+h)-3)-(4x-3)}{h} & & & \\begin{array}{l}\\text{Substitute} \\, f^{\\prime}(x+h)=4(x+h)-3 \\, \\text{and} \\\\ f^{\\prime}(x)=4x-3. \\end{array} \\\\ & =\\underset{h\\to 0}{\\lim}4 & & & \\text{Simplify.} \\\\ & =4 & & & \\text{Take the limit.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738187249\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169738187258\">Find [latex]f''(x)[\/latex] for [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=\"q1238970\">Hint<\/span><\/p>\n<div id=\"q1238970\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738099392\">We found [latex]f^{\\prime}(x)=2x[\/latex] in a previous checkpoint. Use the definition to find the derivative of [latex]f^{\\prime}(x)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738099362\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738099362\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738099362\">[latex]f''(x)=2[\/latex]<\/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=1432&amp;end=1510&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\/3.2TheDerivativeAsAFunction1432to1510_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 id=\"fs-id1169738099443\" class=\"textbook exercises\">\n<h3>Example: Finding Acceleration<\/h3>\n<p id=\"fs-id1169738099452\">The position of a particle along a coordinate axis at time [latex]t[\/latex] (in seconds) is given by [latex]s(t)=3t^2-4t+1[\/latex] (in meters). Find the function that describes its acceleration at time [latex]t[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738219009\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738219009\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738219009\">Since [latex]v(t)=s^{\\prime}(t)[\/latex] and [latex]a(t)=v^{\\prime}(t)=s''(t)[\/latex], we begin by finding the derivative of [latex]s(t)[\/latex]:<\/p>\n<div id=\"fs-id1169737911373\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll}s^{\\prime}(t) & =\\underset{h\\to 0}{\\lim}\\frac{s(t+h)-s(t)}{h} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{3(t+h)^2-4(t+h)+1-(3t^2-4t+1)}{h} \\\\ & =6t-4 \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737950766\">Next,<\/p>\n<div id=\"fs-id1169737950769\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} s''(t) & =\\underset{h\\to 0}{\\lim}\\frac{s^{\\prime}(t+h)-s^{\\prime}(t)}{h} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{6(t+h)-4-(6t-4)}{h} \\\\ & =6 \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738185118\">Thus, [latex]a=6 \\, \\text{m\/s}^2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding Acceleration.<\/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=1513&amp;end=1713&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=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" 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\/3.2TheDerivativeAsAFunction1513to1713_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 id=\"fs-id1169738185145\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169738185153\">For [latex]s(t)=t^3[\/latex], find [latex]a(t)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737935177\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737935177\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737935177\">[latex]a(t)=6t[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1169737935205\">Use the previous examples as a guide.<\/p>\n<\/div>\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-334\">\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":9,"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 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