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Question
\[\frac{1}{a x^2 + bx + c}\]
Solution
\[\frac{d}{dx}\left( \frac{1}{a x^2 + bx + c} \right)\]
\[ = \frac{d}{dx} \left( a x^2 + bx + c \right)^{- 1} \]
\[ = \left( - 1 \right) \left( a x^2 + bx + c \right)^{- 2} \frac{d}{dx}\left( a x^2 + bx + c \right) (\text{ Using the chain rule })\]
\[ = \left( - 1 \right) \left( a x^2 + bx + c \right)^{- 2} \left( 2ax + b \right)\]
\[ = \frac{- \left( 2ax + b \right)}{\left( a x^2 + bx + c \right)^2}\]
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