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Question
\[\frac{2x + 3}{x - 2}\]
Solution
\[ \frac{d}{dx}\left( f(x) \right) = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{\frac{2\left( x + h \right) + 3}{x + h - 2} - \frac{2x + 3}{x - 2}}{h}\]
\[ = \lim_{h \to 0} \frac{\left( 2x + 2h + 3 \right)\left( x - 2 \right) - \left( x + h - 2 \right)\left( 2x + 3 \right)}{h\left( x + h - 2 \right)\left( x - 2 \right)}\]
\[ = \lim_{h \to 0} \frac{2 x^2 + 2xh + 3x - 4x - 4h - 6 - 2 x^2 - 2xh + 4x - 3x - 3h + 6}{h\left( x + h - 2 \right)\left( x - 2 \right)}\]
\[ = \lim_{h \to 0} \frac{- 7h}{h\left( x + h - 2 \right)\left( x - 2 \right)}\]
\[ = \lim_{h \to 0} \frac{- 7}{\left( x + h - 2 \right)\left( x - 2 \right)}\]
\[ = \frac{- 7}{\left( x - 2 \right)\left( x - 2 \right)}\]
\[ = \frac{- 7}{\left( x - 2 \right)^2}\]
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