Question

Find the derivative of the following functions from the first principals w.r.t to $x$.
$\sec 3x$

Answer 1

Applying the first principle to find the derivative of $\sec 3x$.

$\underset{h\to 0}{lim}\frac{\sec 3\left(x+h\right)-\sec 3x}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{1}{\cos 3\left(x+h\right)}-\frac{1}{\cos 3x}}{h}$

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

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

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

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

$=\underset{h\to 0}{lim}\frac{\left(1-\cos 3h\right)}{h\cos 3\left(x+h\right)}+\frac{\sin 3x}{\cos 3x}\times \underset{h\to 0}{lim}\frac{\sin 3h}{h}\times \underset{h\to 0}{lim}\frac{1}{\cos 3\left(x+h\right)}$

$=0+\frac{\sin 3x}{\cos 3x}\times 3\times \frac{1}{\cos 3x}$

$=3\tan 3x\sec 3x$