Derivatives Of Trigonometric Functions

In this section we will learn the formulae for the derivatives of trigonometric functions, and their derivations.

I. Derivative Of sin(x):

Based on the definition of derivatives explained , the derivative of $\sin x$ is given by:

$\dfrac{d }{dx}(\sin x) = \lim\limits_{\Delta x \rightarrow 0} \dfrac{\sin (x + \Delta x) - \sin x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{\sin x \cos \Delta x + \sin \Delta x \cos x - \sin x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{\sin x (\cos \Delta x - 1) + \sin \Delta x \cos x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{\sin \Delta x \cos x - \sin x (1 - \cos \Delta x)}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{\sin \Delta x}{\Delta x} \cos x - \lim\limits_{\Delta x \rightarrow 0} \sin x \dfrac{(1 - \cos \Delta x)}{\Delta x}$

$= \cos x \cdot \lim\limits_{\Delta x \rightarrow 0} \dfrac{\sin \Delta x}{\Delta x} - \sin x \cdot \lim\limits_{\Delta x \rightarrow 0} \dfrac{(1 - \cos \Delta x)}{\Delta x}$

$= 1\cdot \cos x - \sin x \cdot 0$

$= \cos x$

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II. Derivative Of cos(x):

$\dfrac{d}{dx} (\cos x) = \lim\limits_{\Delta x \rightarrow 0} \dfrac{\cos (x + \Delta x) - \cos x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{\cos x \cdot \cos \Delta x - \sin x \cdot \sin \Delta x - \cos x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{\cos x (\cos \Delta x - 1) - \sin x \cdot \sin \Delta x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{- \sin x \cdot \sin \Delta x - \cos x (1 - \cos \Delta x) }{\Delta x}$

$= - \sin x \cdot \lim\limits_{\Delta x \rightarrow 0} \dfrac{\sin \Delta x}{\Delta x} - \cos x \cdot \lim\limits_{\Delta x \rightarrow 0} \dfrac{ (1 - \cos \Delta x) }{\Delta x}$

$= -\sin x \cdot 1 - \cos x \cdot 0$

$= - \sin x$

III. Derivative of tan(x):

$\dfrac{d}{dx}(\tan x)$

$= \dfrac{d}{dx} \left[\dfrac{\sin x}{\cos x}\right]$

$= \dfrac{\cos x \dfrac{d}{dx} (\sin x) - \sin x \dfrac{d}{dx} (\cos x)}{\cos^2 x}$

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$= \dfrac{\cos x \cdot \cos x - \sin x (-\sin x)}{\cos^2 x}$ (using the )

$= \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x}$

$= \dfrac{1}{\cos^2 x}$

$= \sec^2 x$

IV. Derivative Of csc(x):

$\dfrac{d}{dx} \csc (x)$

$= \dfrac{d}{dx} \left[\dfrac{1}{\sin x}\right] = \dfrac{d}{dx} [\sin x]^{-1}$

$= \dfrac{d}{d (\sin x)} [\sin x]^{-1} \cdot \dfrac{d}{dx} \sin x $

$= - [\sin x]^{-2} \cdot \cos x$

$= - \dfrac{\cos x}{\sin^2 x}$

$= - \cot x \cdot \csc x$

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V. Derivative Of sec(x):

$\dfrac{d}{dx}sec(x)$

$= \dfrac{d}{dx} \left[\dfrac{1}{\cos x}\right] = \dfrac{d}{dx} [\cos x]^{-1}$

$= \dfrac{d}{d (\cos x)} [\cos x]^{-1} \cdot \dfrac{d}{dx} \cos x$

$= - [\cos x]^{-2} \cdot (-\sin x)$

$= \dfrac{\sin x}{\cos^2 x} = \tan x \cdot \sec x$

VI. Derivative Of cot(x):

$\dfrac{d}{dx}cot(x)$

$= \dfrac{d}{dx} \left[\dfrac{1}{\tan x}\right] = \dfrac{d}{dx} [\tan x]^{-1}$

$= \dfrac{d}{d (\tan x)} [\tan x]^{-1} \cdot \dfrac{d}{dx} \tan x$

$= - [\tan x]^{-2} \cdot \sec^2 x$

$= -\dfrac{\sec^2 x}{\tan^2 x}$

$= - \csc^2 x$