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