Differentiation of Trigonometric Functions
Differentiation of Trigonometric Functions
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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*}
Example 3:
$$ y=\tan (ax+b),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y &=\tan (ax+b)\\
\frac{dy}{dx} &=\frac{d}{dx}\tan (ax+b)\\\\
\textup{Let}\; u &= ax+b\\\\
\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}(ax+b)\\
\frac{dy}{dx} &=\sec^2 (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} &=\sec^2 (ax+b) (a(1)+0)\\
\frac{dy}{dx} &=\sec^2 (ax+b) (a)\\
\frac{dy}{dx} &=a\sec^2 (ax+b) \\
\end{align*}
Example 4:
$$ y=\cot (ax+b),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y &=\cot (ax+b)\\
\frac{dy}{dx} &=\frac{d}{dx}\cot (ax+b)\\\\
\textup{Let}\; u &= ax+b\\\\
\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}(ax+b)\\
\frac{dy}{dx} &=-\textup{cosec}^2 (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} &=-\textup{cosec}^2 (ax+b) (a(1)+0)\\
\frac{dy}{dx} &=-\textup{cosec}^2 (ax+b) (a)\\
\frac{dy}{dx} &=-a\textup{cosec}^2 (ax+b) \\
\end{align*}
Example 5:
$$ y=\sec (ax+b),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y &=\sec (ax+b)\\
\frac{dy}{dx} &=\frac{d}{dx}\sec (ax+b)\\\\
\textup{Let}\; u &= ax+b\\\\
\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}(ax+b)\\
\frac{dy}{dx} &=\sec (ax+b) \tan (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} &=\sec (ax+b) \tan (ax+b)(a(1)+0)\\
\frac{dy}{dx} &=\sec (ax+b) \tan (ax+b)(a)\\
\frac{dy}{dx} &=a\sec (ax+b) \tan (ax+b)\\
\end{align*}
Example 6:
$$ y=\textup{cosec} (ax+b),\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$
Solution:
\begin{align*}
y &=\textup{cosec} (ax+b)\\
\frac{dy}{dx} &=\frac{d}{dx}\textup{cosec} (ax+b)\\\\
\textup{Let}\; u &= ax+b\\\\
\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}(ax+b)\\
\frac{dy}{dx} &=-\textup{cosec} (ax+b) \cot (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} &=-\textup{cosec} (ax+b) \cot (ax+b)(a(1)+0)\\
\frac{dy}{dx} &=-\textup{cosec} (ax+b) \cot (ax+b)(a)\\
\frac{dy}{dx} &=-a\textup{cosec} (ax+b) \cot (ax+b)\\
\end{align*}