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