$x$ এর সাপেক্ষে $\sec 2 x$ ফাংশনটির অন্তরক সহগ নিচের কোনটি ?

Solve: মনে করি, $\mathrm{f}(x)=\sec 2 x$.$\therefore \mathrm{f}(x+\mathrm{h})=\sec 2(x+\mathrm{h})=\sec (2 \mathrm{x}+2 \mathrm{~h})$

অন্তরক সহগের সংজ্ঞা হতে পাই,

$\begin{aligned} & \frac{d}{d x}\{\mathrm{f}(x)\}=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \therefore \quad & \frac{d}{d x}(\sec 2 x)=\lim _{h \rightarrow 0} \frac{\sec (2 x+2 h)-\sec 2 x}{h} \\ = & \lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{1}{\cos (2 x+2 h)}-\frac{1}{\cos 2 x}\right]^{-} \\ = & \lim _{h \rightarrow 0} \frac{\cos 2 x-\cos (2 x+2 h)}{h \cos (2 x+2 h) \cos 2 x} \end{aligned}$

$\begin{array}{l} =\lim _{h \rightarrow 0} \frac{2 \sin \frac{2 x+2 x+2 h}{2} \sin \frac{2 x+2 h-2 x}{2}}{h \cos (2 x+2 h) \cos 2 x} \\ =2 \lim _{h \rightarrow 0} \frac{\sin (2 x+h)}{\cos (2 x+2 h) \cos 2 x} \times \lim _{h \rightarrow 0} \frac{\sin h}{h} \\ =2 \frac{\sin (2 x+0)}{\cos (2 x+0) \cos 2 x} \times 1 \\ =\frac{2 \sin 2 x}{\cos 2 x \cos 2 x}=2 \tan 2 x \sec 2 x \end{array}$