Find the derivative of the function $\tan 2 x$ , by using the first principle w.r to x

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Detailed Solution

Let $f (x) = Tan 2 x$

By the first principle $f (x + h) = Tan 2 (x + h)$

$f^{'} (x) = {Lt}_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$= \underset{h \to 0}{Lt} \frac{Tan 2 (x + h) - Tan 2 x}{h}$

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

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

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

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

$∴ \sin A \cos B - \cos A \sin B = \sin (A - B)$

$=_{h \to 0}^{l t} {\frac{\sin 2 h}{h}}_{h \to 0}^{l t} \frac{1}{\cos (2 x + 2 h) \cos 2 x}$

$f^{'} (x) = 2 \cdot \frac{1}{\cos^{2} 2 x} = 2 \cdot \sec^{2} 2 x$

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