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प्रश्न
\[\sqrt{2 x^2 + 1}\]
उत्तर
\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2 \left( x + h \right)^2 + 1} - \sqrt{2 x^2 + 1}}{h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2 x^2 + 2 h^2 + 4xh + 1} - \sqrt{2 x^2 + 1}}{h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2 x^2 + 2 h^2 + 4xh + 1} - \sqrt{2 x^2 + 1}}{h} \times \frac{\sqrt{2 x^2 + 2 h^2 + 4xh + 1} + \sqrt{2 x^2 + 1}}{\sqrt{2 x^2 + 2 h^2 + 4xh + 1} + \sqrt{2 x^2 + 1}}\]
\[ = \lim_{h \to 0} \frac{2 x^2 + 2 h^2 + 4xh + 1 - 2 x^2 - 1}{h\left( \sqrt{2 x^2 + 2 h^2 + 4xh + 1} + \sqrt{2 x^2 + 1} \right)}\]
\[ = \lim_{h \to 0} \frac{h\left( 2h + 4x \right)}{h\left( \sqrt{2 x^2 + 2 h^2 + 4xh + 1} + \sqrt{2 x^2 + 1} \right)}\]
\[ = \lim_{h \to 0} \frac{\left( 2h + 4x \right)}{\left( \sqrt{2 x^2 + 2 h^2 + 4xh + 1} + \sqrt{2 x^2 + 1} \right)}\]
\[ = \frac{4x}{\sqrt{2 x^2 + 1} + \sqrt{2 x^2 + 1}}\]
\[ = \frac{4x}{2\sqrt{2 x^2 + 1}}\]
\[ = \frac{2x}{\sqrt{2 x^2 + 1}}\]
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