Differentiate from first principle:
$(i i) t a n (2 x + 1)$

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Solution

$f (x) = t a n (2 x + 1)$

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:

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{t a n [2 (x + h) + 1] - t a n (2 x + 1)}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{\begin{matrix} s i n (2 x + 2 h + 1) c o s (2 x + 1) - c o s (2 x + 2 h + 1) s i n (2 x + 1) \end{matrix}}{h c o s (2 x + 2 h + 1) c o s (2 x + 1)}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{s i n (2 x + 2 h + 1 - 2 x - 1)}{h c o s (2 x + 2 h + 1) c o s (2 x + 1)}$

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

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

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

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

$\Rightarrow f^{'} (x) = 2 s e c^{2} (2 x + 1)$

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