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