Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# √ Tan X - Mathematics

$\sqrt{\tan x}$

#### Solution

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

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 4.4 | Page 26