Find the derivative of the function $\sec 3x$, by using the first principle w.r to x

 

let  $f\left(x\right)=\sec 3x;f(x+h)=\mathrm{Sec}3(x+h)$

By the first principle

$f^{\mathrm{\prime }}\left(x\right)=\underset{h\to 0}{Lt}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{Lt}\frac{\sec (3x+3h)-\sec 3x}{h}$

$=\underset{h\to 0}{Lt}\frac{\cos 3x-\cos (3x+3h)}{h\cos 3x\cos (3x+3h)}$

$=\underset{h\to 0}{Lt}\frac{-2\sin \left(\frac{3x+3x+3h}{2}\right)\sin \left(\frac{3x-3x-3h}{2}\right)}{h\cos 3x\cos (3x+3h)}$

$=\underset{h\to 0}{Lt}\frac{-2\sin \left(\frac{6x+3h}{2}\right)\sin \left(\frac{-3h}{2}\right)}{h\cos 3x\cos (3x+3h)}$

$=\underset{h\to 0}{Lt}\frac{-2\sin \left(\frac{6x+3h}{2}\right)}{\cos 3x}.\underset{h\to 0}{Lt}\frac{-\sin \left(\frac{3h}{2}\right)}{\frac{3}{2}h}\left(\frac{3}{2}\right).\underset{h\to 0}{Lt}\frac{1}{\cos (3x+3h)}$

$f^1\left(x\right)=+2\left(\frac{3}{2}\right)\frac{\sin 3x}{\cos 3x}\cdot 1\cdot \frac{1}{\cos 3x}=3\mathrm{Tan}3x\cdot \sec 3x$