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Question
If \[y = \log\frac{x^2 + x + 1}{x^2 - x + 1} + \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{\sqrt{3} x}{1 - x^2} \right), \text{ find } \frac{dy}{dx} .\] ?
Solution
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)}\frac{d}{dx}\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right) + \frac{2}{\sqrt{3}}\left\{ \frac{1}{1 + \left( \frac{\sqrt{3x}}{1 - x^2} \right)^2} \right\}\frac{d}{dx}\left( \frac{\sqrt{3}x}{1 - x^2} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \left( \frac{x^2 - x + 1}{x^2 + x + 1} \right)\left( \frac{\left( x^2 - x + 1 \right)\frac{d}{dx}\left( x^2 + x + 1 \right) - \left( x^2 + x + 1 \right)\frac{d}{dx}\left( x^2 - x + 1 \right)}{\left( x^2 - x + 1 \right)^2} \right) + \frac{2}{\sqrt{3}}\left\{ \frac{\left( 1 - x^2 \right)^2}{1 + x^4 - 2 x^2 + 3 x^2} \right\} \left\{ \frac{\left( 1 - x^2 \right)\frac{d}{dx}\left( \sqrt{3}x \right) - \sqrt{3}x\frac{d}{dx}\left( 1 - x^2 \right)}{\left( 1 - x^2 \right)^2} \right\}\]
\[ \Rightarrow \frac{dy}{dx} = \left( \frac{1}{x^2 + x + 1} \right)\left\{ \frac{\left( x^2 - x + 1 \right)\left( 2x + 1 \right) - \left( x^2 + x + 1 \right)\left( 2x - 1 \right)}{\left( x^2 - x + 1 \right)} \right\} + \frac{2}{\sqrt{3}}\left\{ \frac{\left( 1 - x^2 \right)^2}{1 + x^2 + x^4} \right\}\left\{ \frac{\left( 1 - x^2 \right)\left( \sqrt{3} \right) - \sqrt{3}x\left( - 2x \right)}{\left( 1 - x^2 \right)^2} \right\}\]
\[ \Rightarrow \frac{dy}{dx} = \left( \frac{2 x^3 - 2 x^2 + 2x + x^2 - x + 1 - 2 x^3 - 2 x^2 - 2x + x^2 + x + 1}{x^4 + 2 x^2 + 1 - x^2} \right) + \frac{2}{\sqrt{3}}\left( \frac{\sqrt{3} - \sqrt{3} x^2 + 2\sqrt{3} x^2}{1 + x^2 + x^4} \right)\]
\[ \Rightarrow \frac{dy}{dx} = \left( \frac{- 2 x^2 + 2}{x^4 + x^2 + 1} \right) + \frac{2\sqrt{3}\left( x^2 + 1 \right)}{\sqrt{3}\left( 1 + x^2 + x^4 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2\left( 1 - x^2 \right)}{\left( x^4 + x^2 + 1 \right)} + \frac{2\left( x^2 + 1 \right)}{1 + x^2 + x^4}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2\left( 1 - x^2 + x^2 + 1 \right)}{1 + x^2 + x^4}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{4}{1 + x^2 + x^4}\]
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f(x) = 3x2 + 6x + 8, x ∈ R
