{"id":231904,"date":"2025-06-11T14:37:06","date_gmt":"2025-06-11T14:37:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231904"},"modified":"2025-06-11T14:37:08","modified_gmt":"2025-06-11T14:37:08","slug":"find-the-35th-derivative-of-the-function-fx-cos2x-at-xi-hint-find-the-pattern-ofthe-nth-derivative-of-the-function-with-n35_","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/find-the-35th-derivative-of-the-function-fx-cos2x-at-xi-hint-find-the-pattern-ofthe-nth-derivative-of-the-function-with-n35_\/","title":{"rendered":"Find the 35th derivative of the function f(x)-cos(2x) at x=I Hint : find the pattern ofthe nth: derivative of the function with n=35_"},"content":{"rendered":"\n

Find the 35th derivative of the function f(x)-cos(2x) at x=I Hint : find the pattern ofthe nth: derivative of the function with n=35_<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the 35th derivative<\/strong> of the function f(x)=cos\u2061(2x)f(x) = \\cos(2x)f(x)=cos(2x)<\/p>\n\n\n\n

at x=ix = ix=i, we first analyze the pattern of the nnnth derivative<\/strong> of the cosine function.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Understand the derivative pattern<\/strong><\/h3>\n\n\n\n

Let\u2019s compute the first few derivatives of f(x)=cos\u2061(2x)f(x) = \\cos(2x)f(x)=cos(2x):<\/p>\n\n\n\n

    \n
  • f(0)(x)=cos\u2061(2x)f^{(0)}(x) = \\cos(2x)f(0)(x)=cos(2x)<\/li>\n\n\n\n
  • f(1)(x)=\u22122sin\u2061(2x)f^{(1)}(x) = -2\\sin(2x)f(1)(x)=\u22122sin(2x)<\/li>\n\n\n\n
  • f(2)(x)=\u22124cos\u2061(2x)f^{(2)}(x) = -4\\cos(2x)f(2)(x)=\u22124cos(2x)<\/li>\n\n\n\n
  • f(3)(x)=8sin\u2061(2x)f^{(3)}(x) = 8\\sin(2x)f(3)(x)=8sin(2x)<\/li>\n\n\n\n
  • f(4)(x)=16cos\u2061(2x)f^{(4)}(x) = 16\\cos(2x)f(4)(x)=16cos(2x)<\/li>\n<\/ul>\n\n\n\n

    A clear pattern emerges:<\/p>\n\n\n\n

      \n
    • The derivative alternates between cosine and sine.<\/li>\n\n\n\n
    • The coefficients follow powers of 2.<\/li>\n\n\n\n
    • The signs follow a cycle: +,\u2212,\u2212,++, -, -, ++,\u2212,\u2212,+ and repeat every 4 steps.<\/li>\n<\/ul>\n\n\n\n

      In general, the nnnth derivative of f(x)=cos\u2061(2x)f(x) = \\cos(2x)f(x)=cos(2x) is: f(n)(x)=2n\u22c5cos\u2061(2x+n\u03c02)f^{(n)}(x) = 2^n \\cdot \\cos\\left(2x + n\\frac{\\pi}{2} \\right)f(n)(x)=2n\u22c5cos(2x+n2\u03c0\u200b)<\/p>\n\n\n\n

      This uses the identity for successive derivatives of cosine with a phase shift.<\/p>\n\n\n\n

      \n\n\n\n

      Step 2: Plug in n=35n = 35n=35 and x=ix = ix=i<\/strong><\/h3>\n\n\n\n

      Using the general form: f(35)(x)=235\u22c5cos\u2061(2x+35\u03c02)f^{(35)}(x) = 2^{35} \\cdot \\cos\\left(2x + 35\\frac{\\pi}{2} \\right)f(35)(x)=235\u22c5cos(2x+352\u03c0\u200b)<\/p>\n\n\n\n

      Substitute x=ix = ix=i: f(35)(i)=235\u22c5cos\u2061(2i+35\u03c02)f^{(35)}(i) = 2^{35} \\cdot \\cos\\left(2i + \\frac{35\\pi}{2} \\right)f(35)(i)=235\u22c5cos(2i+235\u03c0\u200b)<\/p>\n\n\n\n

      Now use the identity: cos\u2061(a+b)=cos\u2061acos\u2061b\u2212sin\u2061asin\u2061b\\cos(a + b) = \\cos a \\cos b – \\sin a \\sin bcos(a+b)=cosacosb\u2212sinasinb<\/p>\n\n\n\n

      Let a=2ia = 2ia=2i, b=35\u03c02b = \\frac{35\\pi}{2}b=235\u03c0\u200b<\/p>\n\n\n\n

      We first find:<\/p>\n\n\n\n

        \n
      • cos\u2061(2i)=cosh\u2061(2)\\cos(2i) = \\cosh(2)cos(2i)=cosh(2) because cos\u2061(iz)=cosh\u2061(z)\\cos(iz) = \\cosh(z)cos(iz)=cosh(z)<\/li>\n\n\n\n
      • sin\u2061(2i)=isinh\u2061(2)\\sin(2i) = i \\sinh(2)sin(2i)=isinh(2) because sin\u2061(iz)=isinh\u2061(z)\\sin(iz) = i \\sinh(z)sin(iz)=isinh(z)<\/li>\n<\/ul>\n\n\n\n

        Now find the angle: 35\u03c02=8\u22c52\u03c0+3\u03c02\u21d2cos\u2061(35\u03c02)=cos\u2061(3\u03c02)=0,sin\u2061(35\u03c02)=\u22121\\frac{35\\pi}{2} = 8\\cdot 2\\pi + \\frac{3\\pi}{2} \\Rightarrow \\cos\\left(\\frac{35\\pi}{2}\\right) = \\cos\\left(\\frac{3\\pi}{2}\\right) = 0, \\quad \\sin\\left(\\frac{35\\pi}{2}\\right) = -1235\u03c0\u200b=8\u22c52\u03c0+23\u03c0\u200b\u21d2cos(235\u03c0\u200b)=cos(23\u03c0\u200b)=0,sin(235\u03c0\u200b)=\u22121<\/p>\n\n\n\n

        So: cos\u2061(2i+35\u03c02)=cos\u2061(2i)\u22c50\u2212sin\u2061(2i)\u22c5(\u22121)=sin\u2061(2i)=isinh\u2061(2)\\cos(2i + \\tfrac{35\\pi}{2}) = \\cos(2i)\\cdot 0 – \\sin(2i)\\cdot (-1) = \\sin(2i) = i\\sinh(2)cos(2i+235\u03c0\u200b)=cos(2i)\u22c50\u2212sin(2i)\u22c5(\u22121)=sin(2i)=isinh(2)<\/p>\n\n\n\n

        Therefore: f(35)(i)=235\u22c5isinh\u2061(2)f^{(35)}(i) = 2^{35} \\cdot i\\sinh(2)f(35)(i)=235\u22c5isinh(2)<\/p>\n\n\n\n

        \n\n\n\n

        Final Answer:<\/strong><\/h3>\n\n\n\n

        f(35)(i)=235\u22c5i\u22c5sinh\u2061(2)\\boxed{f^{(35)}(i) = 2^{35} \\cdot i \\cdot \\sinh(2)}f(35)(i)=235\u22c5i\u22c5sinh(2)\u200b<\/p>\n\n\n\n

        \n\n\n\n

        Explanation<\/strong><\/h3>\n\n\n\n

        To compute the 35th derivative of f(x)=cos\u2061(2x)f(x) = \\cos(2x)f(x)=cos(2x) at x=ix = ix=i, we begin by recognizing the periodic and structured behavior of trigonometric derivatives. The function cos\u2061(2x)\\cos(2x)cos(2x) alternates between cosine and sine on each differentiation, with coefficients increasing as powers of 2. Specifically, the nnnth derivative follows the pattern: f(n)(x)=2n\u22c5cos\u2061(2x+n\u03c02)f^{(n)}(x) = 2^n \\cdot \\cos\\left(2x + n\\frac{\\pi}{2}\\right)f(n)(x)=2n\u22c5cos(2x+n2\u03c0\u200b)<\/p>\n\n\n\n

        This result comes from the rotational symmetry of sine and cosine derivatives. Since cos\u2061(2x)\u2192\u22122sin\u2061(2x)\u2192\u22124cos\u2061(2x)\u21928sin\u2061(2x)\u2192\u22ef\\cos(2x) \\rightarrow -2\\sin(2x) \\rightarrow -4\\cos(2x) \\rightarrow 8\\sin(2x) \\rightarrow \\cdotscos(2x)\u2192\u22122sin(2x)\u2192\u22124cos(2x)\u21928sin(2x)\u2192\u22ef, the nnnth derivative exhibits a predictable rotational shift in phase by \u03c02\\frac{\\pi}{2}2\u03c0\u200b per differentiation.<\/p>\n\n\n\n

        Substituting n=35n = 35n=35, we get: f(35)(x)=235\u22c5cos\u2061(2x+35\u03c02)f^{(35)}(x) = 2^{35} \\cdot \\cos\\left(2x + \\frac{35\\pi}{2}\\right)f(35)(x)=235\u22c5cos(2x+235\u03c0\u200b)<\/p>\n\n\n\n

        Next, we evaluate this at x=ix = ix=i, which is a complex number. Using Euler\u2019s formulas and identities involving complex arguments, we rewrite the cosine of a complex angle: cos\u2061(2i+35\u03c02)=cos\u2061(2i)cos\u2061(35\u03c02)\u2212sin\u2061(2i)sin\u2061(35\u03c02)\\cos(2i + \\tfrac{35\\pi}{2}) = \\cos(2i)\\cos\\left(\\tfrac{35\\pi}{2}\\right) – \\sin(2i)\\sin\\left(\\tfrac{35\\pi}{2}\\right)cos(2i+235\u03c0\u200b)=cos(2i)cos(235\u03c0\u200b)\u2212sin(2i)sin(235\u03c0\u200b)<\/p>\n\n\n\n

        Knowing cos\u2061(2i)=cosh\u2061(2)\\cos(2i) = \\cosh(2)cos(2i)=cosh(2), sin\u2061(2i)=isinh\u2061(2)\\sin(2i) = i\\sinh(2)sin(2i)=isinh(2), and trigonometric values like cos\u2061(35\u03c02)=0\\cos\\left(\\tfrac{35\\pi}{2}\\right) = 0cos(235\u03c0\u200b)=0, sin\u2061(35\u03c02)=\u22121\\sin\\left(\\tfrac{35\\pi}{2}\\right) = -1sin(235\u03c0\u200b)=\u22121, we find: f(35)(i)=235\u22c5i\u22c5sinh\u2061(2)f^{(35)}(i) = 2^{35} \\cdot i \\cdot \\sinh(2)f(35)(i)=235\u22c5i\u22c5sinh(2)<\/p>\n\n\n\n

        This expression is the exact value of the 35th derivative at x=ix = ix=i.<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Find the 35th derivative of the function f(x)-cos(2x) at x=I Hint : find the pattern ofthe nth: derivative of the function with n=35_ The Correct Answer and Explanation is: To find the 35th derivative of the function f(x)=cos\u2061(2x)f(x) = \\cos(2x)f(x)=cos(2x) at x=ix = ix=i, we first analyze the pattern of the nnnth derivative of the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-231904","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231904","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=231904"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231904\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231904"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231904"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231904"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}