Question

Find the derivative of the following by using method of first principle tan(2x+3)​

Answer 1

The derivative of tan(2x+3) using method of first principle is as follows.

Given,

f(x) = tan (2x + 3)

f(x + h) = tan [ 2 (x + h) + 3 ]

$f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}\\\\=\lim_{h \to 0} \dfrac{tan[2(x+h)+3]-tan(2x+3)}{h}$

$= \lim_{h \to 0} \dfrac{tan[(2x+3)+2h] - tan (2x+3)}{h}\\\\= \lim_{h \to 0} \dfrac{tan[(2x+3)+2h]}{1-tan(2x+3)tan2h}-\dfrac{tan (2x+3)}{h}\\\\= \lim_{h \to 0} \dfrac{tan(2x+3)+tan2h-tan(2x+3)(1-tan2htanx)}{\frac{1-tan(2x+3)tan2h}{h}}\\\\= \lim_{h \to 0} \dfrac{tan2h+tan^2(2x+3)tan2h}{h(1-tan(2x+3)tan2h}\\\\= \lim_{h \to 0} \dfrac{tan2h}{h} \times \lim_{h \to 0} \dfrac{1+tan^2(2x+3)}{1-tan(2x+3)tan2h}\\\\= \lim_{h \to 0} \dfrac{sin2h}{cos2h}2 \times \dfrac{tan2h}{2h} \times sec^2(2x+3)$

$\therefore f'(x) = 2sec^2(2x+3)$