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Question
\[\frac{x + 2}{3x + 5}\]
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Solution
\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\frac{x + h + 2}{3\left( x + h \right) + 5} - \frac{x + 2}{3x + 5}}{h}\]
\[ = \lim_{h \to 0} \frac{\frac{x + h + 2}{3x + 3h + 5} - \frac{x + 2}{3x + 5}}{h}\]
\[ = \lim_{h \to 0} \frac{\left( x + h + 2 \right)\left( 3x + 5 \right) - \left( 3x + 3h + 5 \right)\left( x + 2 \right)}{h\left( 3x + 3h + 5 \right)\left( 3x + 5 \right)}\]
\[ = \lim_{h \to 0} \frac{3 x^2 + 3xh + 6x + 5x + 5h + 10 - 3 x^2 - 3xh - 5x - 6x - 6h - 10}{h\left( 3x + 3h + 5 \right)\left( 3x + 5 \right)}\]
\[ = \lim_{h \to 0} \frac{- h}{h\left( 3x + 3h + 5 \right)\left( 3x + 5 \right)}\]
\[ = \lim_{h \to 0} \frac{- 1}{\left( 3x + 3h + 5 \right)\left( 3x + 5 \right)}\]
\[ = \frac{- 1}{\left( 3x + 5 \right)^2}\]
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