Differentiate from first principle:
$(i) {tan}^{2} x$

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Solution

Given:
$f (x) = {tan}^{2} x$

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:

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

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

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

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

$\Rightarrow f^{'} (x) = lim h \to 0 (\begin{matrix} t a n (x + h) + t a n x \end{matrix}) (\begin{matrix} 1 + tan (x + h) t a n x \end{matrix}) \times \frac{t a n h}{h}$

$\Rightarrow f^{'} (x) = (\begin{matrix} t a n (x + 0) + t a n x \end{matrix}) (1 + t a n (x + 0) t a n x) \times 1 [∵ lim x \to 0 \frac{t a n x}{x} = 1]$

$\Rightarrow f^{'} (x) = (2 t a n x) (1 + t a n^{2} x)$

$\Rightarrow f^{'} (x) = 2 t a n x s e c^{2} x$

Therefore, the derivative of $t a n^{2} x$ is $2 t a n x s e c^{2} x$

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