Question

$\frac{d}{d x} (sec x) =$

Answer: $sec x tan x$

\begin{aligned} Let f (x) & = & sec x, then \\ \frac{d f (x)}{d x} & = & lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = & lim_{h \to 0} \frac{sec (x + h) - sec x}{h} \\ = & lim_{h \to 0} \frac{\frac{1}{cos (x + h)} - \frac{1}{cos x}}{h} (∵ sec x = \frac{1}{cos x}) \\ = & lim_{h \to 0} \frac{\frac{cos (x) - cos (x + h)}{cos (x + h) cos x}}{h} \\ = & lim_{h \to 0} \frac{2 sin (\frac{x + x + h}{2}) sin (\frac{x + h - x}{2})}{h cos (x + h) cos x} (∵ cos C - cos D = 2 sin (\frac{C + D}{2}) sin (\frac{D - C}{2})) \\ = & lim_{h \to 0} \frac{2 sin (\frac{2 x + h}{2}) sin (\frac{h}{2})}{h cos (x + h) cos x} \\ = & lim_{h \to 0} \frac{sin (x + \frac{h}{2})}{cos (x + h) cos x} \cdot \frac{sin (\frac{h}{2})}{\frac{h}{2}} \\ = & lim_{h \to 0} \frac{sin (x + \frac{h}{2})}{cos (x + h) cos x} \cdot 1 (∵ lim_{x \to 0} \frac{sin x}{x} = 1) \\ = & \frac{sin (x + \frac{0}{2})}{cos (x + 0) cos x} \\ = & \frac{sin (x + 0)}{cos x cos x} \\ = & \frac{1}{cos x} \cdot \frac{sin x}{cos x} \\ = & sec x tan x \cdot \end{aligned}

So, correct answer is option-3, $\to sec x tan x$