Differentiation of Trigonometric Functions

Example 1:

$$ y=\sin \left(\frac{x}{a}\right),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=\sin \left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\frac{d}{dx}\sin \left(\frac{x}{a}\right)\\\\ \textup{Let}\; u &= \frac{x}{a}\\\\ \frac{dy}{dx} &=\frac{d}{dx} \sin u\\ \frac{dy}{dx} &=\frac{d}{du} \sin u \frac{du}{dx}\\\\ \because\frac{d}{dx}\sin x &=\cos x\\\\ \therefore\frac{dy}{dx} &=\cos u \frac{d}{dx}\left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\cos \left(\frac{x}{a}\right) \left(\frac{1}{a} \frac{d}{dx}x \right)\\\\ \because\frac{d}{dx} x &=1 \\\\ \therefore \frac{dy}{dx} &=\cos \left(\frac{x}{a}\right) \left(\frac{1}{a}(1)\right)\\ \frac{dy}{dx} &=\cos \left(\frac{x}{a}\right) \left(\frac{1}{a}\right)\\ \frac{dy}{dx} &=\frac{1}{a} \cos \left(\frac{x}{a}\right) \\ \end{align*}

Example 2:

$$ y=\cos \left(\frac{x}{a}\right),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=\cos \left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\frac{d}{dx}\cos \left(\frac{x}{a}\right)\\\\ \textup{Let}\; u &= \frac{x}{a}\\\\ \frac{dy}{dx} &=\frac{d}{dx} \cos u\\ \frac{dy}{dx} &=\frac{d}{du} \cos u \frac{du}{dx}\\\\ \because\frac{d}{dx}\cos x &=-\sin x\\\\ \therefore\frac{dy}{dx} &=-\sin u \frac{d}{dx}\left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=-\sin \left(\frac{x}{a}\right) \left(\frac{1}{a} \frac{d}{dx}x \right)\\\\ \because\frac{d}{dx} x &=1 \\\\ \therefore \frac{dy}{dx} &=-\sin \left(\frac{x}{a}\right) \left(\frac{1}{a}(1)\right)\\ \frac{dy}{dx} &=-\sin \left(\frac{x}{a}\right) \left(\frac{1}{a}\right)\\ \frac{dy}{dx} &=-\frac{1}{a}\sin \left(\frac{x}{a}\right) \\ \end{align*}

Example 3:

$$ y=\tan \left(\frac{x}{a}\right),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=\tan \left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\frac{d}{dx}\tan \left(\frac{x}{a}\right)\\\\ \textup{Let}\; u &= \frac{x}{a}\\\\ \frac{dy}{dx} &=\frac{d}{dx} \tan u\\ \frac{dy}{dx} &=\frac{d}{du} \tan u \frac{du}{dx}\\\\ \because\frac{d}{dx}\tan x &=\sec^2 x\\\\ \therefore\frac{dy}{dx} &=\sec^2 u \frac{d}{dx}\left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\sec^2 \left(\frac{x}{a}\right) \left(\frac{1}{a} \frac{d}{dx}x \right)\\\\ \because\frac{d}{dx} x &=1 \\\\ \therefore \frac{dy}{dx} &=\sec^2 \left(\frac{x}{a}\right) \left(\frac{1}{a}(1)\right)\\ \frac{dy}{dx} &=\sec^2 \left(\frac{x}{a}\right) \left(\frac{1}{a}\right)\\ \frac{dy}{dx} &=\frac{1}{a}\sec^2 \left(\frac{x}{a}\right) \\ \end{align*}

Example 4:

$$ y=\cot \left(\frac{x}{a}\right),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=\cot \left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\frac{d}{dx}\cot \left(\frac{x}{a}\right)\\\\ \textup{Let}\; u &= \frac{x}{a}\\\\ \frac{dy}{dx} &=\frac{d}{dx} \cot u\\ \frac{dy}{dx} &=\frac{d}{du} \cot u \frac{du}{dx}\\\\ \because\frac{d}{dx}\cot x &=-\textup{cosec}^2 x\\\\ \therefore\frac{dy}{dx} &=-\textup{cosec}^2 u \frac{d}{dx}\left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=-\textup{cosec}^2 \left(\frac{x}{a}\right)\left(\frac{1}{a} \frac{d}{dx}x \right)\\\\ \because\frac{d}{dx} x &=1 \\\\ \therefore \frac{dy}{dx} &=-\textup{cosec}^2 \left(\frac{x}{a}\right) \left(\frac{1}{a}(1)\right)\\ \frac{dy}{dx} &=-\textup{cosec}^2 \left(\frac{x}{a}\right) \left(\frac{1}{a}\right)\\ \frac{dy}{dx} &=-\frac{1}{a}\textup{cosec}^2 \left(\frac{x}{a}\right) \\ \end{align*}

Example 5:

$$ y=\sec \left(\frac{x}{a}\right),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=\sec \left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\frac{d}{dx}\sec \left(\frac{x}{a}\right)\\\\ \textup{Let}\; u &= \frac{x}{a}\\\\ \frac{dy}{dx} &=\frac{d}{dx} \sec u\\ \frac{dy}{dx} &=\frac{d}{du} \sec u \frac{du}{dx}\\\\ \because\frac{d}{dx}\sec x &=\sec x \tan x\\\\ \therefore\frac{dy}{dx} &=\sec u \tan u \frac{d}{dx}\left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\sec \left(\frac{x}{a}\right) \tan \left(\frac{x}{a}\right) \left(\frac{1}{a} \frac{d}{dx}x \right)\\\\ \because\frac{d}{dx} x &=1 \\\\ \therefore \frac{dy}{dx} &=\sec \left(\frac{x}{a}\right) \tan \left(\frac{x}{a}\right) \left(\frac{1}{a}(1)\right)\\ \frac{dy}{dx} &=\sec \left(\frac{x}{a}\right) \tan \left(\frac{x}{a}\right) \left(\frac{1}{a}\right)\\ \frac{dy}{dx} &=\frac{1}{a}\sec \left(\frac{x}{a}\right) \tan \left(\frac{x}{a}\right)\\ \end{align*}

Example 6:

$$ y=\textup{cosec} \left(\frac{x}{a}\right),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=\textup{cosec} \left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=\frac{d}{dx}\textup{cosec} \left(\frac{x}{a}\right)\\\\ \textup{Let}\; u &= \frac{x}{a}\\\\ \frac{dy}{dx} &=\frac{d}{dx} \textup{cosec} u\\ \frac{dy}{dx} &=\frac{d}{du} \textup{cosec} u \frac{du}{dx}\\\\ \because\frac{d}{dx}\textup{cosec} x &=-\textup{cosec} x \cot x\\\\ \therefore\frac{dy}{dx} &=-\textup{cosec} u \cot u \frac{d}{dx}\left(\frac{x}{a}\right)\\ \frac{dy}{dx} &=-\textup{cosec} \left(\frac{x}{a}\right) \cot \left(\frac{x}{a}\right) \left(\frac{1}{a} \frac{d}{dx}x \right)\\\\ \because\frac{d}{dx} x &=1 \\\\ \therefore \frac{dy}{dx} &=-\textup{cosec} \left(\frac{x}{a}\right) \cot \left(\frac{x}{a}\right) \left(\frac{1}{a}(1)\right)\\ \frac{dy}{dx} &=-\textup{cosec} \left(\frac{x}{a}\right) \cot \left(\frac{x}{a}\right) \left(\frac{1}{a}\right)\\ \frac{dy}{dx} &=-\frac{1}{a}\textup{cosec} \left(\frac{x}{a}\right) \cot \left(\frac{x}{a}\right)\\ \end{align*}