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Question
Differentiate \[\log \left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)\] ?
Solution
\[\text{Let }y = \log\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)\]
Differentiate with respect of x we get,
\[\frac{d y}{d x} = \frac{d}{dx}\left[ \log\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right) \right]\]
\[ = \frac{1}{\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)}\frac{d}{dx}\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right) \left[ \text{Using chain rule and quotient rule} \right]\]
\[ = \left( \frac{x^2 - x + 1}{x^2 + x + 1} \right)\left[ \frac{\left( x^2 - x + 1 \right)\frac{d}{dx}\left( x^2 + x + 1 \right) - \left( x^2 + x + 1 \right)\frac{d}{dx}\left( x^2 - x + 1 \right)}{\left( x^2 - x + 1 \right)^2} \right]\]
\[ = \left( \frac{x^2 - x + 1}{x^2 + x + 1} \right)\left[ \frac{\left( x^2 - x + 1 \right)\left( 2x + 1 \right) - \left( x^2 + x + 1 \right)\left( 2x - 1 \right)}{\left( x^2 - x + 1 \right)^2} \right]\]
\[ = \left( \frac{x^2 - x + 1}{x^2 + x + 1} \right)\left[ \frac{2 x^3 - 2 x^2 + 2x + x^2 - x + 1 - 2 x^3 - 2 x^2 - 2x + x^2 + x + 1}{\left( x^2 - x + 1 \right)^2} \right]\]
\[ = \frac{- 4 x^2 + 2 x^2 + 2}{\left( x^2 + x + 1 \right)\left( x^2 - x + 1 \right)}\]
\[ = \frac{- 4 x^2 + 2 x^2 + 2}{\left( x^2 + 1 \right)^2 - \left( x \right)^2}\]
\[ = \frac{- 2\left( x^2 - 1 \right)}{x^4 + 1 + 2 x^2 - x^2}\]
\[ = \frac{- 2\left( x^2 - 1 \right)}{x^4 + x^2 + 1}\]
\[So, \frac{d}{dx}\left\{ \log\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right) \right\} = \frac{- 2\left( x^2 - 1 \right)}{x^4 + x^2 + 1}\]
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