{"id":8606,"date":"2024-01-02T00:46:51","date_gmt":"2024-01-01T19:16:51","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=8606"},"modified":"2024-01-02T00:46:57","modified_gmt":"2024-01-01T19:16:57","slug":"differentiation-of-trigonometric-functions","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/differentiation-of-trigonometric-functions\/","title":{"rendered":"Differentiation of trigonometric functions"},"content":{"rendered":"\n

Differentiating trigonometric functions is a fundamental concept in calculus. Here’s a quick guide to the derivatives of the basic trigonometric functions. Assume that x<\/em> is a variable and all functions are differentiable.<\/p>\n\n\n\n

Differentiation of Trigonometric Functions<\/h2>\n\n\n\n

Important formula to find the derivative<\/strong>
$\\frac {d}{dx} f(x) =\\displaystyle \\lim_{h \\to 0} \\frac {f (x+h) – f(x)}{h}$
$\\displaystyle \\lim_{x \\to 0} \\frac {sinx}{x}=1$<\/p>\n\n\n\n

Differentiation of Sine Function<\/strong>
$\\frac {d}{dx} (sin x) = cos x $
Proof of sin derivative<\/strong>
$\\frac {d}{dx} (sin x) = \\displaystyle \\lim_{h \\to 0} \\frac {sin (x+h) – sin x}{h}= \\displaystyle \\lim_{h \\to 0} \\frac {2cos (\\frac {2x+h}{2}) sin \\frac {h}{2}}{h}$
$=\\displaystyle \\lim_{h \\to 0} cos (x + \\frac {h}{2}) \\displaystyle \\lim_{h \\to 0} \\frac {sin \\frac {h}{2}}{\\frac {h}{2}} = cos x$
as $\\displaystyle \\lim_{x \\to 0} \\frac {sinx}{x}=1$<\/p>\n\n\n\n

Differentiation<\/strong> of Cos Function<\/strong>
$\\frac {d}{dx} (cos x) = -sin x $
Proof<\/strong>
$\\frac {d}{dx} (cos x) = \\displaystyle \\lim_{h \\to 0} \\frac {cos (x+h) – cos x}{h}= \\displaystyle \\lim_{h \\to 0} -\\frac {2sin (\\frac {2x+h}{2}) sin \\frac {h}{2}}{h}$
$=-\\displaystyle \\lim_{h \\to 0} sin (x + \\frac {h}{2}) \\displaystyle \\lim_{h \\to 0} \\frac {sin \\frac {h}{2}}{\\frac {h}{2}} = -sin x$
as $\\displaystyle \\lim_{x \\to 0} \\frac {sinx}{x}=1$<\/p>\n\n\n\n

Differentiation of Tan Function<\/strong>
$\\frac {d}{dx} (tan x) = sec^2x $
Proof<\/strong>
$\\frac {d}{dx} (tan x) = \\displaystyle \\lim_{h \\to 0} \\frac {tan (x+h) – tan x}{h}= \\displaystyle \\lim_{h \\to 0} \\frac {1}{h}[\\frac {sin(x+h)}{cos(x+h) }- \\frac {sin(x)}{cos(x)}]$
$=\\displaystyle \\lim_{h \\to 0} \\frac {1}{h}\\frac {sin(x+h)cos (x) – sin(x) cos (x+h)}{cos (x+h) cos x}$
$=\\displaystyle \\lim_{h \\to 0} \\frac {1}{h} \\frac {sin (x+h -x)}{cos (x+h) cos x} $
$=\\displaystyle \\lim_{h \\to 0} \\frac {sin h}{h} \\displaystyle \\lim_{h \\to 0} \\frac {1}{cos (x+h) cos x}= sec^2 x$
as $\\displaystyle \\lim_{x \\to 0} \\frac {sinx}{x}=1$<\/p>\n\n\n\n

Alternatively, since $ \\tan(x) = \\frac{\\sin(x)}{\\cos(x)} $, you can use the quotient rule to derive this.<\/p>\n\n\n\n

$\\frac {d}{dx} [f(x)\/g(x)]=\\frac {g(x) \\frac {d}{dx} f(x) – f(x) \\frac {d}{dx} g(x)}{[g(x)]^2} $<\/p>\n\n\n\n

$\\frac {d}{dx} [sin(x)\/cos(x)]=\\frac {cos(x) cos(x) – sin(x) (-sin(x))}{[cos(x)]^2} = \\frac {1}{[cos(x)]^2}= sec^2 x$<\/p>\n\n\n\n

Differentiation<\/strong> of Cot Function<\/strong>
$\\frac {d}{dx} (cot x) = -cosec^2x $
Proof<\/strong><\/p>\n\n\n\n

Now $ \\cot(x) = \\frac{\\cos(x)}{\\sin(x)} $, and the quotient rule applies.<\/p>\n\n\n\n

$\\frac {d}{dx} [f(x)\/g(x)]=\\frac {g(x) \\frac {d}{dx} f(x) – f(x) \\frac {d}{dx} g(x)}{[g(x)]^2} $<\/p>\n\n\n\n

$\\frac {d}{dx} [cos(x)\/sin(x)]=\\frac {sin(x) (-sin(x)) – cos(x) cos(x)}{[sin(x)]^2} = \\frac {-1}{[sin(x)]^2}= -cosec^2 x$<\/p>\n\n\n\n

Differentiation<\/strong> of Secant Function<\/strong>
$\\frac {d}{dx} (sec x) = sec x . tan x $<\/p>\n\n\n\n

Proof<\/strong><\/p>\n\n\n\n

Now $\\sec (x) = \\frac {1}{cos(x)}$ , and the quotient rule applies.<\/p>\n\n\n\n

$\\frac {d}{dx} [f(x)\/g(x)]=\\frac {g(x) \\frac {d}{dx} f(x) – f(x) \\frac {d}{dx} g(x)}{[g(x)]^2} $<\/p>\n\n\n\n

$\\frac {d}{dx} [1\/cos(x)]=\\frac {0 + sin(x)}{[cos(x)]^2} = sec x . tan x $
<\/p>\n\n\n\n

Differentiation<\/strong> of Cosecant Function<\/strong>
$\\frac {d}{dx} (cosec x) = -cosec x . cot x$<\/p>\n\n\n\n

Proof<\/strong><\/p>\n\n\n\n

Now $cosec (x) = \\frac {1}{sin(x)}$ , and the quotient rule applies.<\/p>\n\n\n\n

$\\frac {d}{dx} [f(x)\/g(x)]=\\frac {g(x) \\frac {d}{dx} f(x) – f(x) \\frac {d}{dx} g(x)}{[g(x)]^2} $<\/p>\n\n\n\n

$\\frac {d}{dx} [1\/sin(x)]=\\frac {0 – cos(x)}{[sin(x)]^2} = -cosec x . cot x$<\/p>\n\n\n\n

Chain Rule for composite Trigonometric Functions<\/h2>\n\n\n\n


When differentiating composite trigonometric functions, the chain rule is often necessary. For example, if you have a function like $ \\sin(2x) $, its derivative is $ 2\\cos(2x) $, obtained by applying the chain rule.<\/p>\n\n\n\n

Examples<\/strong><\/p>\n\n\n\n

    \n
  1. Differentiating $ \\sin(3x^2) $
    \\[ \\frac{d}{dx}[\\sin(3x^2)] = \\cos(3x^2) \\cdot 6x \\]
    Here, the chain rule is applied: first, differentiate ( \\sin(u) ) with respect to ( u ), and then multiply by the derivative of ( u = 3x^2 ) with respect to ( x ).<\/li>\n\n\n\n
  2. Differentiating $ \\tan(x^3) $
    \\[ \\frac{d}{dx}[\\tan(x^3)] = \\sec^2(x^3) \\cdot 3x^2 \\]
    Again, the chain rule is used.<\/li>\n<\/ol>\n\n\n\n

    Higher Order Derivatives<\/h2>\n\n\n\n

    For higher-order derivatives, trigonometric functions exhibit cyclic patterns. For instance, the derivatives of sine and cosine functions repeat every four derivatives.<\/p>\n\n\n\n

    $\\frac {d}{dx} sin (x) = cos (x)$
    $\\frac {d^2}{dx^2} cos (x) = -sin (x)$
    $\\frac {d^3}{dx^3} (-sin (x)) = -cos (x)$
    $\\frac {d^4}{dx^4} (-cos (x)) = sin (x)$<\/p>\n\n\n\n

    Differentiation of Inverse Trigonometric Functions<\/h2>\n\n\n\n
    \"\"<\/a><\/figure>\n","protected":false},"excerpt":{"rendered":"

    Differentiating trigonometric functions is a fundamental concept in calculus. Here’s a quick guide to the derivatives of the basic trigonometric functions. Assume that x is a variable and all functions are differentiable. Differentiation of Trigonometric Functions Important formula to find the derivative$\\frac {d}{dx} f(x) =\\displaystyle \\lim_{h \\to 0} \\frac {f (x+h) – f(x)}{h}$$\\displaystyle \\lim_{x \\to […]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","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":"","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":"set","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":[498],"tags":[],"class_list":["post-8606","post","type-post","status-publish","format-standard","hentry","category-maths"],"yoast_head":"\nDifferentiation of trigonometric functions - physicscatalyst's Blog<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physicscatalyst.com\/article\/differentiation-of-trigonometric-functions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Differentiation of trigonometric functions - physicscatalyst's Blog\" \/>\n<meta property=\"og:description\" content=\"Differentiating trigonometric functions is a fundamental concept in calculus. 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