Find the derivative of the function $\sec 3 x$ , by using the first principle w.r to x

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Detailed Solution

let $f (x) = \sec 3 x; f (x + h) = Sec 3 (x + h)$

By the first principle

$f^{'} (x) = \underset{h \to 0}{L t} \frac{f (x + h) - f (x)}{h}$

$= \underset{h \to 0}{L t} \frac{\sec (3 x + 3 h) - \sec 3 x}{h}$

$= \underset{h \to 0}{L t} \frac{\cos 3 x - \cos (3 x + 3 h)}{h \cos 3 x \cos (3 x + 3 h)}$

$= \underset{h \to 0}{L t} \frac{- 2 \sin (\frac{3 x + 3 x + 3 h}{2}) \sin (\frac{3 x - 3 x - 3 h}{2})}{h \cos 3 x \cos (3 x + 3 h)}$

$= \underset{h \to 0}{L t} \frac{- 2 \sin (\frac{6 x + 3 h}{2}) \sin (\frac{- 3 h}{2})}{h \cos 3 x \cos (3 x + 3 h)}$

$= \underset{h \to 0}{L t} \frac{- 2 \sin (\frac{6 x + 3 h}{2})}{\cos 3 x} . \underset{h \to 0}{L t} \frac{- \sin (\frac{3 h}{2})}{\frac{3}{2} h} (\frac{3}{2}) . \underset{h \to 0}{L t} \frac{1}{\cos (3 x + 3 h)}$

$f^{1} (x) = + 2 (\frac{3}{2}) \frac{\sin 3 x}{\cos 3 x} \cdot 1 \cdot \frac{1}{\cos 3 x} = 3 Tan 3 x \cdot \sec 3 x$

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