Differentiation of Trigonometric Functions
Differentiation of Trigonometric Functions
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Derivative of tan x
\begin{align*}
\frac{d}{dx}\tan x &=\frac{d}{dx} \frac{\sin x}{\cos x} \\\\
\because\;\; \frac{d}{dx} \frac{u}{v} &= \frac{v \frac{du}{dx}-u \frac{dv}{dx}}{v^2} \\\\
\therefore\;\; \frac{d}{dx}\tan x &=\frac{\cos x \frac{d}{dx}\sin x - \sin x \frac{d}{dx}\cos x}{(\cos x)^2} \\
&=\frac{\cos x (\cos x) - \sin x(-\sin x)}{\cos^2 x} \\
&=\frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\
&=\frac{1}{\cos^2 x} \\
\therefore\frac{d}{dx}\tan x &=\sec^2 x \\
\end{align*}
Derivative of cot x
\begin{align*}
\frac{d}{dx}\cot x &=\frac{d}{dx} \frac{\cos x}{\sin x} \\\\
\because\;\; \frac{d}{dx} \frac{u}{v} &= \frac{v \frac{du}{dx}-u \frac{dv}{dx}}{v^2} \\\\
\therefore\;\; \frac{d}{dx}\cot x &=\frac{\sin x \frac{d}{dx}\cos x - \cos x \frac{d}{dx}\sin x}{(\sin x)^2} \\
&=\frac{\sin x (-\sin x) - \cos x(\cos x)}{\cos^2 x} \\
&=\frac{-\cos^2 x - \sin^2 x}{\sin^2 x} \\
&=-\frac{\cos^2 x + \sin^2 x}{\sin^2 x} \\
&=-\frac{1}{\sin^2 x} \\
\therefore\frac{d}{dx}\cot x &=-\textup{cosec}^2 x \\
\end{align*}
Derivative of sec x
\begin{align*}
\frac{d}{dx}\sec x &=\frac{d}{dx} \frac{1}{\cos x} \\
&= \frac{d}{dx}(\cos x)^{-1}\\\\
\textup{Let}\;\;\; u &= \cos x \\\\
\therefore \frac{d}{dx}\sec x &=\frac{d}{dx} u^{-1}\\
&=\frac{d}{du} u^{-1} \frac{du}{dx} \\\\
\because\;\; \frac{d}{dx} x^n &=nx^{n-1}\\\\
\therefore \frac{d}{dx}\sec x &= -1u^{-1-1} \frac{d}{dx} \cos x\\
&= -1u^{-2} (-\sin x)\\
&= -\frac{1}{u^2} (-\sin x)\\
&= -\frac{1}{\cos^2 x} (-\sin x) \\
&= \frac{\sin x}{\cos^2 x} \\
&= \left(\frac{1}{\cos x} \right) \left(\frac{\sin x}{\cos x}\right)\\
\therefore\frac{d}{dx}\sec x &=\sec x \tan x\\
\end{align*}
Derivative of cosec x
\begin{align*}
\frac{d}{dx}\textup{cosec} x &=\frac{d}{dx} \frac{1}{\sin x} \\
&= \frac{d}{dx}(\sin x)^{-1}\\\\
\textup{Let}\;\;\; u &= \sin x \\\\
\therefore \frac{d}{dx}\textup{cosec} x &=\frac{d}{dx} u^{-1}\\
&=\frac{d}{du} u^{-1} \frac{du}{dx} \\\\
\because\;\; \frac{d}{dx} x^n &=nx^{n-1}\\\\
\therefore \frac{d}{dx}\textup{cosec} x &= -1u^{-1-1} \frac{d}{dx} \sin x\\
&= -1u^{-2} (\cos x)\\
&= -\frac{1}{u^2} (\cos x)\\
&= -\frac{1}{\sin^2 x} (\cos x) \\
&= -\frac{\cos x}{\sin^2 x} \\
&= -\left(\frac{1}{\sin x} \right) \left(\frac{\cos x}{\sin x}\right)\\
\therefore\frac{d}{dx}\textup{cosec} x &= -\textup{cosec} x \cot x\\
\end{align*}