Question

Differentiate $\tan^{2} x$ using first principle.

Accepted Answer

We have to find derivative of $\tan^{2} x$ using first principle of differentiation

$\frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$= \lim_{h \to 0} \frac{\tan^{2} (x + h) - \tan^{2} x}{h}$

$= \lim_{h \to 0} \frac{[\tan (x + h) + \tan x] [\tan (x + h) - \tan x]}{h}$

$= \lim_{h \to 0} \frac{[\frac{\sin (x + h)}{\cos (x + h)} + \frac{\sin x}{\cos x}] [\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}]}{h}$

$= \lim_{h \to 0} \frac{[\sin (x + h) \cos x + \cos (x + h) \sin x] [\sin (x + h) \cos x - \cos (x + h) \sin x]}{h \cos^{2} x \cos^{2} (x + h)}$

$= \lim_{h \to 0} \frac{[\sin (2 x + h)] [\sin h]}{h \cos^{2} x \cos^{2} (x + h)}$

$= \frac{1}{\cos^{2} x} \lim_{h \to 0} \sin (2 x + h) \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{\cos^{2} (x + h)}$

$= \frac{1}{\cos^{2} x} \sin (2 x) (1) \frac{1}{\cos^{2} x}$

$= \frac{1}{\cos^{2} x} 2 \sin x \cos x \frac{1}{\cos^{2} x}$

$= 2 \times \frac{\sin x}{\cos x} \times \frac{1}{\cos^{2} x}$

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

R D Sharma Solutions for Chapter: Derivatives, Exercise 2: EXERCISE 30.2

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Derivatives, Exercise 2: EXERCISE 30.2

Questions from R D Sharma Solutions for Chapter: Derivatives, Exercise 2: EXERCISE 30.2 with Hints & Solutions

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