Derivative of Trigonometric Functions

Other derivative involves trigonometric. This is quite different from the Algebraic. In this page, we again deal only with the first derivative. To make it more sense, the functions were given below.

Trigonometric Function Derivatives:

$i.) \frac{d}{dx}sin {\color{Red} u}=cos {\color{Red} u}\frac{d{\color{Red} u}}{dx}$

$ii.)\frac{d}{dx}cos {\color{Red} u}=-sin {\color{Red} u}\frac{d{\color{Red} u}}{dx}$

$iii.)\frac{d}{dx}tan {\color{Red} u}=sec^{2} {\color{Red} u}\frac{d{\color{Red} u}}{dx}$

$iv.)\frac{d}{dx}cot {\color{Red} u}=-csc^{2} {\color{Red} u}\frac{d{\color{Red} u}}{dx}$

$v.)\frac{d}{dx}sec {\color{Red} u}=sec {\color{Red} u}tan{\color{Red} u}\frac{d{\color{Red} u}}{dx}$

$vi.)\frac{d}{dx}csc{\color{Red} u}=-csc {\color{Red} u}cot{\color{Red} u}\frac{d{\color{Red} u}}{dx}$

Illustrative Examples:

Find the first derivative of the given trigonometric functions. This is deriving it's basics.

$1.) y= sin{\color{Red} 2x}$ , # here our u = 2x

$y{}'=\frac{dy}{dx}=\frac{d}{dx}sin{\color{Red} 2x}$ , # We are finding the derivative of sin 2x here

${y}'=\frac{dy}{dx} = cos{\color{Red} 2x} \frac{d}{dx} ({\color{Red} 2x})$ $;\frac{d}{dx}({\color{Red}2x})={\color{Red}2}$ $;see(i)$

${y}'=\frac{dy}{dx} = 2cos{\color{Red} 2x}$ , answer

$2.) y= cos{\color{Red} 5x},u=5x$ , # here our u = 5x

$y{}'= \frac{dy}{dx} = \frac{d}{dx}cos{\color{Red} 5x}$ , # this is to find the derivative of cos 5x

$y{}'=\frac{dy}{dx} =-sin{\color{Red} 5x}\frac{d}{dx}({\color{Red} 5x})$ $;\frac{d}{dx}({\color{Red} 5x})={\color{Red} 5}$ $;see(ii)$

$y{}'=\frac{dy}{dx} =-5sin{\color{Red} 5x}$ , answer

$3.)y=tan{\color{Red} 3x}$ ; # here our u = 3x

${y}'=\frac{dy}{dx}=sec^{2}{\color{Red} 3x}\frac{d}{dx}{\color{Red} 3x}$ $:\frac{d}{dx}{\color{Red} 3x}={\color{Red} 3}$ , # this is to find the derivative of $sec^{2}{\color{Red} 3x}$

$y{}'=\frac{dy}{dx}=3sec^{2}{\color{Red} 3x}$ , answer

Examples Problems:

Find the trigonometric derivatives of the given functions.

Note:

${y}'=\frac{dy}{dx}$

$1.)y=cot{\color{Red} 2x}$ $;u={\color{Red} 2x},u{}'={\color{Red} 2}$

$y{}'=\frac{dy}{dx}=-csc^{2}{\color{Red} 2x}\frac{d}{dx}{\color{Red} 2x}$

$y{}'=-2csc^{2}{\color{Red} 2x}$ , answer

$2.) x =2cos{\color{Red} 4t}$

$x{}'=\frac{dx}{dt}=2(-sin{\color{Red} 4t}\frac{d}{dt}{\color{Red} 4t})$

$x{}'=\frac{dx}{dt}=-2sin4t({\color{Red} 4})$

$x{}'=\frac{dx}{dt}=-8sin{\color{Red} 4t}$ , answer

$3.)u=3csc({\color{Red} 2-3v})$

${u}'{}=\frac{du}{dv}[3csc({\color{Red} 2-3v})]$

${u}'=-3csc({\color{Red} 2-3}v)cot({\color{Red}{\color{Red}2-3v}})\frac{d}{dv}({\color{Red}2-3v})$

${u}'=-3csc({\color{Red} 2-3}v)cot({\color{Red}{\color{Red}2-3v}})({\color{Red}-3})$

${u}'=9csc({\color{Red} 2-3}v)cot({\color{Red}{\color{Red}2-3v}})$ , answer

$4.) y=3sec{\color{Red} 2x}$

${y}'=\frac{dy}{dx}=\frac{d}{dx}(3sec{\color{Red} 2x})$

${y}'=3sec{\color{Red}2x}tan{\color{Red}2x}\frac{d}{dx}({\color{Red}3x})$

${y}'=3sec{\color{Red} 2x}tan{\color{Red} 2x}({\color{Red} 3})$

${y}'=9sec{\color{Red} 2x}tan{\color{Red} 2x}$ , answer

$5.){y}=cos^{3}{\color{Red}3x}$

${y}'=\frac{dy}{dx}=3cos^{2}{\color{Red}3x}(-sin{\color{Red} 3x})\frac{d}{dx}({\color{Red} 3x})$

${y}'=3cos^{2}{\color{Red}3x}(-sin{\color{Red} 3x})({\color{Red} 3})$

${y}'=9cos^{2}{\color{Red}3x}sin{\color{Red}3x}$ , answer

Practice Test:

Find the trigonometric derivatives of the given functions.

$1.) y=sin6x$

${y}'=cos6x\frac{d}{dx}(6x)$

${y}'=$ click for the answer

$2.) y=cot 4x$

$y'=sec^{2}4x\frac{d}{dx}(4x)$

${y}'=$ click for the answer

$3.)y=3cos2x$

$y'=3[-sin2x\frac{d}{dx}(2x)]$

$y'=3[-sin2x(2)]$

${y}'=$ click for the answer

Simple Quiz:

Find the first derivative of the given trigonometric functions.

$1.) y=-sin{\color{Red} 2x}$

$2.)y=-2cos{\color{Red} 3x}$

$3.)x=tan{\color{Red} 4\theta }$

$4.)w =cot{\color{Red} 3z}$

$5.)v =-5cos{\color{Red}3u}$

Leave us comment if any typographical errors appears

Contact us:

Facebook, Twitter, Instagram, Tumblr,

Related links: link 1, link 2

"Leave your Comments"

DMG

Derivative of Trigonometric Functions

Posted by : Allan_Dell on Friday, April 17, 2020 | 9:38 AM

Popular posts

DMG

Derivative of Trigonometric Functions

Posted by : Allan_Dell on Friday, April 17, 2020 | 9:38 AM

Popular posts

Posted by : Allan_Dell on Friday, April 17, 2020 | 9:38 AM