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Question:

If $y=\log \frac{x^{2}+x+1}{x^{2}-x+1}+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$, find $\frac{d y}{d x}$

Solution:

Here, $y=\log \frac{x^{2}+x+1}{x^{2}-x+1}+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$

Differentiating it with respect to $x$ using chain and quotient rule,

$\frac{d y}{d x}=\frac{d}{d x} \log \frac{x^{2}+x+1}{x^{2}-x+1}+\frac{2}{\sqrt{3}} \frac{d}{d x} \tan ^{-1}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$

$\frac{d y}{d x}=\frac{1}{\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)} \frac{d}{d x}\left(\frac{x^{2}+x+1}{x^{2}-x+1}\right)+\frac{2}{\sqrt{3}}\left\{\frac{1}{1+\left(\frac{\sqrt{3} x}{1-x^{2}}\right)}\right\} \frac{d}{d x}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}$

$=\left(\frac{\mathrm{x}^{2}-\mathrm{x}+1}{\mathrm{x}^{2}+\mathrm{x}+1}\right)\left(\frac{\left(\mathrm{x}^{2}-\mathrm{x}+1\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+\mathrm{x}+1\right)-\left(\mathrm{x}^{2}+\mathrm{x}+1\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-\mathrm{x}+1\right)}{\left(\mathrm{x}^{2}-\mathrm{x}+1\right)^{2}}\right)$

$++\frac{2}{\sqrt{3}}\left\{\frac{(1-x)^{2}}{1+x^{4}-2 x^{2}+3 x^{2}}\right\}\left\{\frac{\left(1-x^{2}\right)^{2} \frac{d}{d x}(\sqrt{3} x)-\sqrt{3} x \frac{d}{d x}(1-x)^{2}}{\left(1-x^{2}\right)^{2}}\right\}$

$\frac{d y}{d x}=\left(\frac{1}{x^{2}+x+1}\right)\left(\frac{\left(x^{2}-x+1\right)(2 x+1)-\left(x^{2}+x+1\right)(2 x-1)}{x^{2}-x+1}\right)$

$+\frac{2}{\sqrt{3}}\left(\frac{\left(1-x^{2}\right)^{2}}{1+x^{2}+x^{4}}\right)\left(\frac{\left(1-x^{2}\right)(\sqrt{3})-\sqrt{3} x(-2 x)}{\left(1-x^{2}\right)^{2}}\right)$

$\frac{d y}{d x}=\left(\frac{2 x^{3}-2 x^{2}+2 x+x^{2}-x+1-2 x^{3}-2 x^{2}+2 x+x^{2}+x+1}{x^{2}-x+1}\right)$

$+\frac{2}{\sqrt{3}}\left(\frac{\sqrt{3}-\sqrt{3} x^{2}+2 \sqrt{3} x^{2}}{1+x^{2}+x^{4}}\right)$

$=\left(\frac{-2 x^{2}+2}{x^{4}+x^{2}+1}\right)+\frac{2 \sqrt{3}\left(x^{2}+1\right)}{\sqrt{3}\left(1+x^{2}+x^{4}\right)}$

$=\left(\frac{2\left(1-x^{2}\right)}{x^{4}+x^{2}+1}\right)+\frac{2\left(x^{2}+1\right)}{\left(1+x^{2}+x^{4}\right)}$

$=\frac{2\left(1-x^{2}+x^{2}+1\right)}{\left(1+x^{2}+x^{4}\right)}$

Hence,

$\frac{d y}{d x}=\frac{4}{\left(1+x^{2}+x^{4}\right)}$

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