Question

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

Answer 1

$\mathrm{We}\mathrm{have},y=\log \left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3x}}{1-x^2}\right)$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\log \left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{2}{\sqrt{3}}\frac{d}{dx}{\tan }^{-1}\left(\frac{\sqrt{3x}}{1-x^2}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{\left({\displaystyle \frac{x^2+x+1}{x^2-x+1}}\right)}\frac{d}{dx}\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{2}{\sqrt{3}}\left\{\frac{1}{1+{\left({\displaystyle \frac{\sqrt{3x}}{1-x^2}}\right)}^2}\right\}\frac{d}{dx}\left(\frac{\sqrt{3}x}{1-x^2}\right) \\ \Rightarrow \frac{dy}{dx}=\left(\frac{x^2-x+1}{x^2+x+1}\right)\left(\frac{\left(x^2-x+1\right){\displaystyle \frac{d}{dx}}\left(x^2+x+1\right)-\left(x^2+x+1\right){\displaystyle \frac{d}{dx}}\left(x^2-x+1\right)}{{\left(x^2-x+1\right)}^2}\right)+\frac{2}{\sqrt{3}}\left\{\frac{{\left(1-x^2\right)}^2}{1+x^4-2x^2+3x^2}\right\}\left\{\frac{\left(1-x^2\right){\displaystyle \frac{d}{dx}}\left(\sqrt{3}x\right)-\sqrt{3}x{\displaystyle \frac{d}{dx}}\left(1-x^2\right)}{{\left(1-x^2\right)}^2}\right\} \\ \Rightarrow \frac{dy}{dx}=\left(\frac{1}{x^2+x+1}\right)\left\{\frac{\left(x^2-x+1\right)\left(2x+1\right)-\left(x^2+x+1\right)\left(2x-1\right)}{\left(x^2-x+1\right)}\right\}+\frac{2}{\sqrt{3}}\left\{\frac{{\left(1-x^2\right)}^2}{1+x^2+x^4}\right\}\left\{\frac{\left(1-x^2\right)\left(\sqrt{3}\right)-\sqrt{3}x\left(-2x\right)}{{\left(1-x^2\right)}^2}\right\} \\ \Rightarrow \frac{dy}{dx}=\left(\frac{2x^3-2x^2+2x+x^2-x+1-2x^3-2x^2-2x+x^2+x+1}{x^4+2x^2+1-x^2}\right)+\frac{2}{\sqrt{3}}\left(\frac{\sqrt{3}-\sqrt{3}{x}^2+2\sqrt{3}{x}^2}{1+x^2+x^4}\right) \\ \Rightarrow \frac{dy}{dx}=\left(\frac{-2x^2+2}{x^4+x^2+1}\right)+\frac{2\sqrt{3}\left(x^2+1\right)}{\sqrt{3}\left(1+x^2+x^4\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{2\left(1-x^2\right)}{\left(x^4+x^2+1\right)}+\frac{2\left(x^2+1\right)}{1+x^2+x^4} \\ \Rightarrow \frac{dy}{dx}=\frac{2\left(1-x^2+x^2+1\right)}{1+x^2+x^4} \\ \Rightarrow \frac{dy}{dx}=\frac{4}{1+x^2+x^4} \end{gathered}$$