Question

Find the derivative of the following functions from the first principals w.r.t to $x$.
$\tan 2x$

Answer 1

We know, first principal is,

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

So with $f\left(x\right)=\tan 2x$ we have

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\tan (2x+2h)-\tan 2x}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{\sin (2x+2h)}{\cos (2x+2h)}-\frac{\sin 2x}{\cos 2x}}{h}$ [ $\tan A=\frac{\sin A}{\cos A}$]

$=\underset{h\to 0}{lim}\frac{1}{h}[\frac{\sin (2x+2h)}{\cos (2x+2h)}-\frac{\sin 2x}{\cos 2x}]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{\sin (2x+2h)\cos 2x-\cos (2x+2h)\sin 2x}{\cos (2x+2h)\cos 2x}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{\sin (2x+2h-2x)}{\cos (2x+2h)\cos 2x}\right]$ [ $\sin (A-B)=\sin A.\cos B-\cos B.\sin A$]

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{\sin \left(2h\right)}{\cos (2x+2h)\cos 2x}\right]$

$=\underset{h\to 0}{lim}\frac{1}{\cos (2x+2h)\cos 2x}.\frac{\sin 2h}{h}$

$=\underset{h\to 0}{lim}\frac{2}{\cos (2x+2h)\cos 2x}.\frac{\sin 2h}{2h}$

$=\underset{h\to 0}{lim}\frac{2}{\cos (2x+2h)\cos 2x}.\underset{h\to 0}{lim}\frac{\sin 2h}{2h}$

We know, $limh\to 0\frac{\sin h}{h}=1\Rightarrow \underset{h\to 0}{lim}\frac{\sin 2h}{2h}=1$

$f^{\prime }\left(x\right)=\frac{2}{\cos 2x.\cos 2x}.1$

$=\frac{2}{{\cos }^{2}2x}$

$=2{\sec }^{2}2x$