# Tan2 X - Mathematics

tan2

#### Solution

$\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}$
$= \lim_{h \to 0} \frac{\tan^2 \left( x + h \right) - \tan^2 x}{h}$
$= \lim_{h \to 0} \frac{\left[ \tan \left( x + h \right) + \tan x \right]\left[ \tan \left( x + h \right) - \tan x \right]}{h}$
$= \lim_{h \to 0} \frac{\left[ \frac{\sin \left( x + h \right)}{\cos \left( x + h \right)} + \frac{\sin x}{\cos x} \right]\left[ \frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x} \right]}{h}$
$= \lim_{h \to 0} \frac{\left[ \sin \left( x + h \right) \cos x + \cos \left( x + h \right) \sin x \right]\left[ \sin \left( x + h \right) \cos x - \cos \left( x + h \right) \sin x \right]}{h \cos^2 x \cos^2 \left( x + h \right)}$
$= \lim_{h \to 0} \frac{\left[ \sin \left( 2x + h \right) \right]\left[ \sin h \right]}{h \cos^2 x \cos^2 \left( x + h \right)}$
$= \frac{1}{\cos^2 x} \lim_{h \to 0} \sin \left( 2x + h \right) \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{\cos^2 \left( x + h \right)}$
$= \frac{1}{\cos^2 x} \sin \left( 2x \right) \left( 1 \right)\frac{1}{\cos^2 x}$
$= \frac{1}{\cos^2 x} 2 \sin x \cos x \frac{1}{\cos^2 x}$
$= 2 \times \frac{\sin x}{\cos x} \times \frac{1}{\cos^2 x}$
$= 2 \tan x \sec^2 x$

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.2 | Q 4.1 | Page 26