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Question
\[\frac{x^2 - x + 1}{x^2 + x + 1}\]
Solution
\[\text{ Let } u = x^2 - x + 1; v = x^2 + x + 1\]
\[\text{ Then }, u' = 2x - 1; v' = 2x + 1\]
\[\text{ By quotient rule },\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{x^2 - x + 1}{x^2 + x + 1} \right) = \frac{\left( x^2 + x + 1 \right)\left( 2x - 1 \right) - \left( x^2 - x + 1 \right)\left( 2x + 1 \right)}{\left( x^2 + x + 1 \right)^2}\]
\[ = \frac{2 x^3 + 2 x^2 + 2x - x^2 - x - 1 - 2 x^3 + 2 x^2 - 2x - x^2 + x - 1}{\left( x^2 + x + 1 \right)^2}\]
\[ = \frac{2 x^2 - 2}{\left( x^2 + x + 1 \right)^2}\]
\[ = \frac{2\left( x^2 - 1 \right)}{\left( x^2 + x + 1 \right)^2}\]
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