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Question
\[\frac{1}{a x^2 + bx + c}\]
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Solution
\[\text{ Let } u = 1; v = a x^2 + bx + c\]
\[\text{ Then }, u' = 0; v' = 2ax + b\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{1}{a x^2 + bx + c} \right) = \frac{\left( a x^2 + bx + c \right)0 - 1\left( 2ax + b \right)}{\left( a x^2 + bx + c \right)^2}\]
\[ = \frac{- \left( 2ax + b \right)}{\left( a x^2 + bx + c \right)^2}\]
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