# Sin √ 2 X - Mathematics

$\sin \sqrt{2x}$

#### Solution

$\text{ Let } f(x) = \sin\sqrt{2x}$
$\text{ Thus, we have }:$
$f(x + h) = \sin\sqrt{2\left( x + h \right)}$
$\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}$
$= \lim_{h \to 0} \frac{\sin \sqrt{2x + 2h} - \sin \sqrt{2x}}{h}$
$\text{ We know }:$
$sin C- sin D=2 sin\left( \frac{C - D}{2} \right)\cos\left( \frac{C + D}{2} \right)$
$= \lim_{h \to 0} \frac{2 \sin\left( \sqrt{2x + 2h} - \sqrt{2x} \right) \cos\left( \sqrt{2x + 2h} - \sqrt{2x} \right)}{h}$
$= \lim_{h \to 0} \frac{2 \times 2 \sin\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right) \cos\left( \frac{\sqrt{2x + 2h} + \sqrt{2x}}{2} \right)}{2h + 2x - 2x}$
$= \lim_{h \to 0} \frac{2 \times 2 \sin\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right) \cos\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right)}{\left( \sqrt{2x + 2h} - \sqrt{2x} \right)\sqrt{2x + 2h} + \sqrt{2x}}$
$= \lim_{h \to 0} \frac{2 \times 2 \sin\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right) \cos\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right)}{2 \times \left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right)\left( \sqrt{2x + 2h} + \sqrt{2x} \right)}$
$= \lim_{h \to 0} \frac{\sin\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right)}{\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right)} \lim_{h \to 0} \frac{2\cos \left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right)}{\sqrt{2x + 2h} + \sqrt{2x}}$
$= 1 \times \frac{2\cos\sqrt{2x}}{2\sqrt{2x}} \left[ \because \lim_{h \to 0} \frac{\sin\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right)}{\left( \frac{\sqrt{2x + 2h} - \sqrt{2x}}{2} \right)} = 1 \right]$
$= \frac{\cos\sqrt{2x}}{\sqrt{2x}}$

Concept: The Concept of Derivative - Algebra of Derivative of Functions
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.2 | Q 5.1 | Page 26