{"id":249,"date":"2023-09-20T22:48:34","date_gmt":"2023-09-20T22:48:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/higher-order-derivatives-of-trig-functions\/"},"modified":"2024-08-05T12:45:44","modified_gmt":"2024-08-05T12:45:44","slug":"derivatives-of-trigonometric-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-of-trigonometric-functions-learn-it-3\/","title":{"raw":"Derivatives of Trigonometric Functions: Learn It 3","rendered":"Derivatives of Trigonometric Functions: Learn It 3"},"content":{"raw":"<h2>Higher-Order Derivatives of Trig Functions<\/h2>\r\n<p id=\"fs-id1169739303785\">The higher-order derivatives of [latex]\\sin x[\/latex] and [latex]\\cos x[\/latex] exhibit a cyclical pattern, making it possible to predict any higher-order derivative of these functions. By understanding this repeating sequence, you can easily compute derivatives beyond the first order.<\/p>\r\n<p>To illustrate, let's calculate the first four derivatives of [latex]y= \\sin x[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}<br \/>\r\n\\text{Start with the function itself:} &amp; y = \\sin x \\\\<br \/>\r\n\\text{The first derivative of } \\sin x \\text{ is:} &amp; \\dfrac{dy}{dx} = \\cos x \\\\<br \/>\r\n\\text{The second derivative becomes:} &amp; \\dfrac{d^2y}{dx^2} = -\\sin x \\\\<br \/>\r\n\\text{Continuing, the third derivative is:} &amp; \\dfrac{d^3y}{dx^3} = -\\cos x \\\\<br \/>\r\n\\text{The fourth derivative brings us back to the starting function:} &amp; \\dfrac{d^4y}{dx^4} = \\sin x<br \/>\r\n\\end{array}[\/latex]<\/p>\r\n<p>This sequence of derivatives demonstrates a pattern that repeats every four derivatives.<\/p>\r\n<ul>\r\n\t<li>[latex]\\sin x[\/latex] leads to [latex]\\cos x[\/latex]\u00a0<\/li>\r\n\t<li>[latex]\\cos x[\/latex] leads to [latex]- \\sin x[\/latex]<\/li>\r\n\t<li>[latex]- \\sin x[\/latex] leads to [latex]- \\cos x[\/latex]<\/li>\r\n\t<li>[latex]- \\cos x[\/latex] leads back to [latex]\\sin x[\/latex]<\/li>\r\n<\/ul>\r\n<p>Understanding this cyclical pattern not only simplifies calculations but also equips us with a systematic approach for determining any higher-order derivative.<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How to: Use the Cyclic Pattern to Determine Higher-Order Derivatives of Sine and Cosine Functions<\/strong><\/p>\r\n<ol>\r\n\t<li>Determine the order of the derivative you need (let's call it [latex]n[\/latex]).<\/li>\r\n\t<li>Calculate the remainder when [latex]n[\/latex] is divided by [latex]4[\/latex]. The remainder determines the position in the cycle:\r\n\r\n<ul>\r\n\t<li><strong>Remainder 0:<\/strong> The derivative returns to the original function.<\/li>\r\n\t<li><strong>Remainder 1:<\/strong> The derivative progresses to the next function in the cycle.<\/li>\r\n\t<li><strong>Remainder 2:<\/strong> The derivative is the negative of the original function.<\/li>\r\n\t<li><strong>Remainder 3:<\/strong> The derivative is the negative of the next function in the cycle.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox proTip\">\r\n<ul>\r\n\t<li>For [latex]\\sin x[\/latex]:\r\n\r\n<ul>\r\n\t<li>If [latex]n[\/latex] is divisible by [latex]4[\/latex] (remainder [latex]0[\/latex]), the [latex]n[\/latex]-th derivative is [latex]\\sin x[\/latex].<\/li>\r\n\t<li>If [latex]n[\/latex]\u00a0divided by [latex]4[\/latex] gives a remainder of [latex]1[\/latex], it is [latex]\\cos x[\/latex].<\/li>\r\n\t<li>If the remainder is [latex]2[\/latex], it is [latex]- \\sin x[\/latex].<\/li>\r\n\t<li>If the remainder is [latex]3[\/latex], it is [latex]- \\cos x[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li>For [latex]\\cos x[\/latex]:\r\n\r\n<ul>\r\n\t<li>Remainder [latex]0[\/latex]: [latex]\\cos x[\/latex]<\/li>\r\n\t<li>Remainder [latex]1[\/latex]: [latex]- \\sin x[\/latex]<\/li>\r\n\t<li>Remainder [latex]2[\/latex]: [latex]- \\cos x[\/latex]<\/li>\r\n\t<li>Remainder [latex]3[\/latex]: [latex]\\sin x[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739293614\">Find [latex]\\dfrac{d^{74}}{dx^{74}}(\\sin x)[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739293655\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739293655\"]<\/p>\r\n<p id=\"fs-id1169739293655\">We can see right away that for the [latex]74[\/latex]th derivative of [latex]\\sin x, \\, 74=4(18)+2[\/latex], so,<\/p>\r\n<div id=\"fs-id1169739293692\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{d^{74}}{dx^{74}}(\\sin x)=\\dfrac{d^{72+2}}{dx^{72+2}}(\\sin x)=\\dfrac{d^2}{dx^2}(\\sin x)=\u2212\\sin x[\/latex].<\/div>\r\n\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]224399[\/ohm_question]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739376135\">A particle moves along a coordinate axis in such a way that its position at time [latex]t[\/latex] is given by [latex]s(t)=2- \\sin t[\/latex].<\/p>\r\n<p>Find [latex]v\\left(\\dfrac{\\pi}{4}\\right)[\/latex]\u00a0 and\u00a0 [latex]a\\left(\\dfrac{\\pi}{4}\\right)[\/latex]. Compare these values and decide whether the particle is speeding up or slowing down.<\/p>\r\n<p id=\"fs-id1169739273581\">[reveal-answer q=\"519394\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"519394\"]<\/p>\r\n<p>First find [latex]v(t)=s^{\\prime}(t)[\/latex]: [latex]v(t)=s^{\\prime}(t)=\u2212\\cos t[\/latex]. Thus, [latex]v\\left(\\dfrac{\\pi}{4}\\right)=-\\dfrac{1}{\\sqrt{2}}[\/latex].<\/p>\r\n<p>Next, find [latex]a(t)=v^{\\prime}(t)[\/latex].<\/p>\r\n<p>Thus, [latex]a(t)=v^{\\prime}(t)= \\sin t[\/latex] and we have [latex]a\\left(\\dfrac{\\pi}{4}\\right)=\\dfrac{1}{\\sqrt{2}}[\/latex].<\/p>\r\n<p>Since [latex]v\\left(\\dfrac{\\pi}{4}\\right)=-\\dfrac{1}{\\sqrt{2}}&lt;0[\/latex] and [latex]a\\left(\\dfrac{\\pi}{4}\\right)=\\dfrac{1}{\\sqrt{2}}&gt;0[\/latex], we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is traveling.<\/p>\r\n<p>Consequently, the particle is slowing down.<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/hvsQJFir7Qw?controls=0&amp;start=1492&amp;end=1573&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.5DerivativesOfTrigonometricFunctions1492to1573_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.5 Derivatives of Trigonometric Functions (edited)\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<h2>Higher-Order Derivatives of Trig Functions<\/h2>\n<p id=\"fs-id1169739303785\">The higher-order derivatives of [latex]\\sin x[\/latex] and [latex]\\cos x[\/latex] exhibit a cyclical pattern, making it possible to predict any higher-order derivative of these functions. By understanding this repeating sequence, you can easily compute derivatives beyond the first order.<\/p>\n<p>To illustrate, let&#8217;s calculate the first four derivatives of [latex]y= \\sin x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}<br \/>  \\text{Start with the function itself:} & y = \\sin x \\\\<br \/>  \\text{The first derivative of } \\sin x \\text{ is:} & \\dfrac{dy}{dx} = \\cos x \\\\<br \/>  \\text{The second derivative becomes:} & \\dfrac{d^2y}{dx^2} = -\\sin x \\\\<br \/>  \\text{Continuing, the third derivative is:} & \\dfrac{d^3y}{dx^3} = -\\cos x \\\\<br \/>  \\text{The fourth derivative brings us back to the starting function:} & \\dfrac{d^4y}{dx^4} = \\sin x<br \/>  \\end{array}[\/latex]<\/p>\n<p>This sequence of derivatives demonstrates a pattern that repeats every four derivatives.<\/p>\n<ul>\n<li>[latex]\\sin x[\/latex] leads to [latex]\\cos x[\/latex]\u00a0<\/li>\n<li>[latex]\\cos x[\/latex] leads to [latex]- \\sin x[\/latex]<\/li>\n<li>[latex]- \\sin x[\/latex] leads to [latex]- \\cos x[\/latex]<\/li>\n<li>[latex]- \\cos x[\/latex] leads back to [latex]\\sin x[\/latex]<\/li>\n<\/ul>\n<p>Understanding this cyclical pattern not only simplifies calculations but also equips us with a systematic approach for determining any higher-order derivative.<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>How to: Use the Cyclic Pattern to Determine Higher-Order Derivatives of Sine and Cosine Functions<\/strong><\/p>\n<ol>\n<li>Determine the order of the derivative you need (let&#8217;s call it [latex]n[\/latex]).<\/li>\n<li>Calculate the remainder when [latex]n[\/latex] is divided by [latex]4[\/latex]. The remainder determines the position in the cycle:\n<ul>\n<li><strong>Remainder 0:<\/strong> The derivative returns to the original function.<\/li>\n<li><strong>Remainder 1:<\/strong> The derivative progresses to the next function in the cycle.<\/li>\n<li><strong>Remainder 2:<\/strong> The derivative is the negative of the original function.<\/li>\n<li><strong>Remainder 3:<\/strong> The derivative is the negative of the next function in the cycle.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox proTip\">\n<ul>\n<li>For [latex]\\sin x[\/latex]:\n<ul>\n<li>If [latex]n[\/latex] is divisible by [latex]4[\/latex] (remainder [latex]0[\/latex]), the [latex]n[\/latex]-th derivative is [latex]\\sin x[\/latex].<\/li>\n<li>If [latex]n[\/latex]\u00a0divided by [latex]4[\/latex] gives a remainder of [latex]1[\/latex], it is [latex]\\cos x[\/latex].<\/li>\n<li>If the remainder is [latex]2[\/latex], it is [latex]- \\sin x[\/latex].<\/li>\n<li>If the remainder is [latex]3[\/latex], it is [latex]- \\cos x[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li>For [latex]\\cos x[\/latex]:\n<ul>\n<li>Remainder [latex]0[\/latex]: [latex]\\cos x[\/latex]<\/li>\n<li>Remainder [latex]1[\/latex]: [latex]- \\sin x[\/latex]<\/li>\n<li>Remainder [latex]2[\/latex]: [latex]- \\cos x[\/latex]<\/li>\n<li>Remainder [latex]3[\/latex]: [latex]\\sin x[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739293614\">Find [latex]\\dfrac{d^{74}}{dx^{74}}(\\sin x)[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739293655\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739293655\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739293655\">We can see right away that for the [latex]74[\/latex]th derivative of [latex]\\sin x, \\, 74=4(18)+2[\/latex], so,<\/p>\n<div id=\"fs-id1169739293692\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{d^{74}}{dx^{74}}(\\sin x)=\\dfrac{d^{72+2}}{dx^{72+2}}(\\sin x)=\\dfrac{d^2}{dx^2}(\\sin x)=\u2212\\sin x[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm224399\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=224399&theme=lumen&iframe_resize_id=ohm224399&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739376135\">A particle moves along a coordinate axis in such a way that its position at time [latex]t[\/latex] is given by [latex]s(t)=2- \\sin t[\/latex].<\/p>\n<p>Find [latex]v\\left(\\dfrac{\\pi}{4}\\right)[\/latex]\u00a0 and\u00a0 [latex]a\\left(\\dfrac{\\pi}{4}\\right)[\/latex]. Compare these values and decide whether the particle is speeding up or slowing down.<\/p>\n<p id=\"fs-id1169739273581\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q519394\">Show Solution<\/button><\/p>\n<div id=\"q519394\" class=\"hidden-answer\" style=\"display: none\">\n<p>First find [latex]v(t)=s^{\\prime}(t)[\/latex]: [latex]v(t)=s^{\\prime}(t)=\u2212\\cos t[\/latex]. Thus, [latex]v\\left(\\dfrac{\\pi}{4}\\right)=-\\dfrac{1}{\\sqrt{2}}[\/latex].<\/p>\n<p>Next, find [latex]a(t)=v^{\\prime}(t)[\/latex].<\/p>\n<p>Thus, [latex]a(t)=v^{\\prime}(t)= \\sin t[\/latex] and we have [latex]a\\left(\\dfrac{\\pi}{4}\\right)=\\dfrac{1}{\\sqrt{2}}[\/latex].<\/p>\n<p>Since [latex]v\\left(\\dfrac{\\pi}{4}\\right)=-\\dfrac{1}{\\sqrt{2}}<0[\/latex] and [latex]a\\left(\\dfrac{\\pi}{4}\\right)=\\dfrac{1}{\\sqrt{2}}>0[\/latex], we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is traveling.<\/p>\n<p>Consequently, the particle is slowing down.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/hvsQJFir7Qw?controls=0&amp;start=1492&amp;end=1573&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.5DerivativesOfTrigonometricFunctions1492to1573_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.5 Derivatives of Trigonometric Functions (edited)&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":6,"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\":\"3.5 Derivatives of Trigonometric Functions (edited)\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":560,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"3.5 Derivatives of Trigonometric Functions (edited)","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\/249"}],"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":12,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/249\/revisions"}],"predecessor-version":[{"id":4516,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/249\/revisions\/4516"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/560"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/249\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=249"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=249"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=249"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=249"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}