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

Tan 2x - Mathematics | Shaalaa.com

Question

tan 2x

Solution

$\frac{d}{d x} (f (x)) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
$= lim_{h \to 0} \frac{\tan (2 x + 2 h) - \tan (2 x)}{h}$
$= lim_{h \to 0} \frac{\frac{s i n (2 x + 2 h)}{\cos (2 x + 2 h)} - \frac{\sin (2 x)}{\cos (2 x)}}{h}$
$= lim_{h \to 0} \frac{s i n (2 x + 2 h) \cos (2 x) - \cos (2 x + 2 h) \sin (2 x)}{h \cos (2 x + 2 h) \cos (2 x)}$
$= lim_{h \to 0} \frac{\sin (2 x + 2 h - 2 x)}{h \cos (2 x + 2 h) \cos (2 x)}$
$= \frac{1}{\cos 2 x} lim_{h \to 0} \frac{\sin (2 h)}{2 h} \times 2 \times lim_{h \to 0} \frac{1}{\cos (2 x + 2 h)}$
$= \frac{1}{\cos 2 x} \times 2 \times \frac{1}{\cos 2 x}$
$= \frac{2}{\cos^{2} (2 x)}$
$= 2 \sec^{2} (2 x)$

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The Concept of Derivative - Algebra of Derivative of Functions

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Q 4.3 Q 4.2 Q 4.4

Chapter 30: Derivatives - Exercise 30.2 [Page 26]

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Chapter 30 Derivatives
Exercise 30.2 | Q 4.3 | Page 26

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Tan 2x - Mathematics

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