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प्रश्न
\[\tan \sqrt{x}\]
उत्तर
\[Let f(x) = \tan\sqrt{x}\]
\[\text{ Thus, we have }: \]
\[(x + h) = \tan\sqrt{x + h}\]
\[\frac{d}{dx}(f(x)) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]
\[ = \lim_{h \to 0} \frac{\tan\sqrt{x + h} - \tan\sqrt{x}}{h}\]
\[ = \lim_{h \to 0} \frac{\sin \left( \sqrt{x + h} - \sqrt{x} \right)}{h \cos\sqrt{x + h} \cos \sqrt{x}} \left[ \because \tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B} \right] \]
\[ = \lim_{h \to 0} \frac{\sin \left( \sqrt{x + h} - \sqrt{x} \right)}{\left( x + h - x \right) \cos\sqrt{x + h} \cos \sqrt{x}} \]
\[ = \lim_{h \to 0} \frac{\sin \left( \sqrt{x + h} - \sqrt{x} \right)}{\left( \sqrt{x + h} - \sqrt{x} \right)\left( \sqrt{x + h} - \sqrt{x} \right)\cos\sqrt{x + h} \cos \sqrt{x}}\]
\[ = \lim_{h \to 0} \frac{\sin \left( \sqrt{x + h} - \sqrt{x} \right)}{\left( \sqrt{x + h} - \sqrt{x} \right)} . \lim_{h \to 0} \frac{1}{\left( \sqrt{x + h} + \sqrt{x} \right)\cos\sqrt{x + h}\cos\sqrt{x}} \left[ \because \lim_{h \to 0} \frac{\sin\left( \sqrt{x + h} - \sqrt{x} \right)}{\sqrt{x + h} - \sqrt{x}} = 1 \right]\]
\[ = 1 \times \frac{1}{2\sqrt{x}\cos\sqrt{x}\cos\sqrt{x}}\]
\[ = \frac{1}{2\sqrt{x}} \sec^2 \sqrt{x}\]
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