Differentiation of Trigonometric Functions

Example 1:

$$ y=\sin (ax+b),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=\sin (ax+b)\\ \frac{dy}{dx} &=\frac{d}{dx}\sin (ax+b)\\\\ \textup{Let}\; u &= (ax+b)\\\\ \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}(ax+b)\\ \frac{dy}{dx} &=\cos (ax+b) \left(a\frac{d}{dx}x+\frac{d}{dx}b \right)\\\\ \because\frac{d}{dx} x &=1 \textup{ and } \frac{d}{dx}b=0\\\\ \therefore \frac{dy}{dx} &=\cos (ax+b) (a(1)+0)\\ \frac{dy}{dx} &=\cos (ax+b) (a)\\ \frac{dy}{dx} &=a\cos (ax+b) \\ \end{align*}

Example 2:

$$ y=\cos (ax+b),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y &=\cos (ax+b)\\ \frac{dy}{dx} &=\frac{d}{dx}\cos (ax+b)\\\\ \textup{Let}\; u &= (ax+b)\\\\ \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}(ax+b)\\ \frac{dy}{dx} &=-\sin (ax+b) \left(a\frac{d}{dx}x+\frac{d}{dx}b \right)\\\\ \because\frac{d}{dx} x &=1 \textup{ and } \frac{d}{dx}b=0\\\\ \therefore \frac{dy}{dx} &=-\sin (ax+b) (a(1)+0)\\ \frac{dy}{dx} &=-\sin (ax+b) (a)\\ \frac{dy}{dx} &=-a\sin (ax+b) \\ \end{align*}