Given function is f(x) = sec√x

By first principle of derivative, we have,

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

\(= \lim\limits_{h \to 0} \frac{sec\sqrt{(x + h)} - sec\sqrt{x}}{h}\)

\(= \lim\limits_{h \to 0} \cfrac{\frac{1}{cos\sqrt{(x + h)}} - \frac{1}{cos\sqrt{x}}}{h}\)

\(= \lim\limits_{h \to 0} \frac{cos\sqrt{x} - cos \sqrt{x + h}}{h\,cos\sqrt{x}\,cos\sqrt{x + h}}\)

\(= \lim\limits_{h \to 0} \cfrac{2\,sin \left(\frac{\sqrt{x} + \sqrt{x + h}}{2}\right)\, sin \left(\frac{\sqrt{x + h} - \sqrt{x}}{2}\right)}{h\,cos \sqrt{x}\,cos \sqrt{x + h}}\)

(∵ cos C - cos D = \(2\,sin \frac{C + D}{2}\) \(sin \frac{D - C}{2}\))

\(=2 \lim\limits_{h \to 0} \cfrac{sin \left(\frac{\sqrt{x} + \sqrt{x + h}}{2}\right)}{cos \sqrt{x}}\) × \(\lim\limits_{h \to 0} \frac{1}{cos\sqrt{x + h}}\) × \(\lim\limits_{h \to 0} \cfrac{sin \left(\frac{\sqrt{x + h} - \sqrt{x}}{2}\right)}{h}\)

\(\left(\because \lim\limits_{h \to 0}\,f(x) y(x) = \left(\lim\limits_{h \to 0} f(x) \right) \left(\lim\limits_{h \to 0} y(x) \right)\right)\)

\(= 2 \cfrac{sin \left(\frac{2\sqrt{x}}{2}\right)}{cos\sqrt{x}}\) × \(\frac{1}{cos\sqrt{x}}\) × \(\lim\limits_{h \to 0} \cfrac{sin\frac{\sqrt{x + h} - \sqrt{x}}{2}}{\frac{\sqrt{x + h} - \sqrt{x}}{2}} \times \cfrac{\frac{\sqrt{x + h} - \sqrt{x}}{2}}{h}\)

(Multiplying both mumerator and denominator by \(\frac{\sqrt{x + h} - \sqrt{x}}{2}\))

= 2 tan √x sec √x × 1 × \(\lim\limits_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{2h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}\)

(∵ \(\lim\limits_{x \to 0} \frac{sin\,x}{x} = 1\) and multiplying both numerator and denominator by \((\sqrt{x + h} + \sqrt{x})\))

= 2 tan √x sec √x × \(\lim\limits_{h \to 0} \frac{x + h - x}{2h(\sqrt{x + h} + \sqrt{x})}\) (∵ (a + b)(a - b) = a² - b²)

= 2 tan √x sec √x × \(\lim\limits_{h \to 0} \frac{h}{2h(\sqrt{x + h} + \sqrt{x})}\)

=  tan √x sec √x × \(\lim\limits_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}\)

\(= \frac{1}{2\sqrt{x}}\,tan \sqrt{x}\,sec \sqrt{x}\) (By taking limit)

∴ \(\frac{d}{dx} sec \sqrt{x} = \frac{1}{2\sqrt{x}}\,tan \sqrt{x}\,sec\sqrt{x}.\)