Differentiate from first principle:
$(i i i) t a n 2 x$

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Solution

Given:

$f (x) = t a n 2 x$

The derivative of a function $f (x)$ is defined as:

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

Putting $f (x)$ in above expression, we get:

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{t a n [2 (x + h)] - t a n (2 x)}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{t a n (2 x + 2 h) - t a n (2 x)}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{\frac{s i n (2 x + 2 h)}{c o s (2 x + 2 h)} - \frac{s i n (2 x)}{c o s (2 x)}}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{\begin{matrix} s i n (2 x + 2 h) c o s (2 x) - c o s (2 x + 2 h) s i n (2 x) \end{matrix}}{h c o s (2 x + 2 h) c o s (2 x)}$

$\Rightarrow f^{'} = lim h \to 0 \frac{s i n (2 x + 2 h - 2 x)}{h c o s (2 x + 2 h) c o s (2 x)}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{s i n (2 h)}{2 h} \times \frac{2}{c o s (2 x + 2 h) c o s (2 x)}$

$\Rightarrow f^{'} (x) = 1 \times \frac{2}{c o s (2 x + 0) c o s (2 x)}$

$[∵ lim h \to 0 \frac{s i n (h)}{h} = 1]$

$\Rightarrow f^{'} (x) = \frac{2}{c o s^{2} (2 x)}$

$\Rightarrow f^{'} (x) = 2 s e c^{2} (2 x)$

Therefore, the derivative of tan 2x is $2 s e c^{2} (2 x)$ .

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