Question

If $y=\frac{1}{3}\log \frac{x+1}{\sqrt{(x^2-x+1)}}+\frac{1}{\sqrt{\left(3\right)}}{\tan }^{-1}\frac{2x-1}{\sqrt{\left(3\right)}},$ then $\frac{dy}{dx}$ at $x=0$?

Answer 1

$y=\frac{1}{3}\log \frac{x+1}{\sqrt{(x^2-x+1)}}+\frac{1}{\sqrt{\left(3\right)}}{\tan }^{-1}\frac{2x-1}{\sqrt{\left(3\right)}}$
$y=\frac{1}{3}\log (x+1)-\frac{1}{6}\log (x^2-x+1)+\frac{1}{\sqrt{\left(3\right)}}{\tan }^{-1}\frac{2x-1}{\sqrt{\left(3\right)}}$
$\frac{dy}{dx}=\frac{1}{3}\frac{1}{x+1}-\frac{1}{6}\frac{2x-1}{x^2-x+1}+\frac{1}{\sqrt{\left(3\right)}}\cdot \frac{1}{1+\frac{(2x-1{)}^2}{3}}\frac{2}{\sqrt{3}}$
$=\frac{1}{6}\frac{2(x^2-x+1)-(x+1)(2x-1))}{x^3+1}+\frac{2}{3}\cdot \frac{3}{3+4x^2-4x+1}=\frac{1}{6}\frac{-3x+3}{x^3+1}+\frac{1}{2(x^2-x+1)}$
$=\frac{1}{2}\frac{1-x}{x^3+1}+\frac{x+1}{2(x+1)(x^2-x+1)}=\frac{1}{2(x^3+1)}\cdot [1-x+x+1]=\frac{1}{x^3+1}\therefore y^{\prime }\left(0\right)=1$