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प्रश्न
Differentiate each of the following from first principle:
\[\sqrt{\sin 2x}\]
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उत्तर
\[\frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{\sin \left( 2x + 2h \right)} - \sqrt{\sin 2x}}{h} \times \frac{\sqrt{\sin \left( 2x + 2h \right)} + \sqrt{\sin 2x}}{\sqrt{\sin \left( 2x + 2h \right)} + \sqrt{\sin 2x}}\]
\[ = \lim_{h \to 0} \frac{\sin \left( 2x + 2h \right) - \sin 2x}{h \left( \sqrt{\sin \left( 2x + 2h \right)} + \sqrt{\sin 2x} \right)}\]
\[\text{ We have }:\]
\[sin C-sin D= 2 cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right)\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( \frac{2x + 2h + 2x}{2} \right) \sin \left( \frac{2x + 2h - 2x}{2} \right)}{h \left( \sqrt{\sin \left( 2x + 2h \right)} + \sqrt{\sin 2x} \right)}\]
\[ = \lim_{h \to 0} \frac{2 \cos \left( 2x + h \right) \sin h}{h \left( \sqrt{\sin \left( 2x + 2h \right)} + \sqrt{\sin 2x} \right)}\]
\[ = \lim_{h \to 0} 2 \cos \left( 2x + h \right) \lim_{h \to 0} \frac{\sin h}{h} \lim_{h \to 0} \frac{1}{\left( \sqrt{\sin \left( 2x + 2h \right)} + \sqrt{\sin 2x} \right)} \]
\[ = 2 \cos 2x \left( 1 \right) \frac{1}{\sqrt{\sin 2x} + \sqrt{\sin 2x}}\]
\[ = \frac{2 \cos 2x}{2\sqrt{\sin 2x}}\]
\[ = \frac{\cos 2x}{\sqrt{\sin 2x}}\]
\[\]
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