If `y=log(x^2+x+1)/(x^2-x+1)+2/(sqrt(3))t a n^(-1)((sqrt(3)x)/(1-x^2)),"f i n d"(dy)/(dx)`

Explanation

We have, `y=log (frac{x^{2}+x+1}{x^{2}-x+1})+frac{2}{sqrt{3}} tan ^{-1}(frac{sqrt{3 x}}{1-x^{2}})`

Differentiating with respect to `x` using chain 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}(frac{sqrt{3 x}}{1-x^{2}})`

`Rightarrow frac{d y}{d x}=frac{1}{(frac{x^{2}+x+1}{x^{2}-x+1})} frac{d}{d x}(frac{x^{2}+x+1}{x^{2}-x+1})+frac{2}{sqrt{3}}{frac{1}{1+(frac{sqrt{3 x}}{1-x^{2}})^{2}}} frac{d}{d x}(frac{sqrt{3} x}{1-x^{2}})`

`Rightarrow frac{d y}{d x}=(frac{x^{2}-x+1}{x^{2}+x+1})(frac{(x^{2}-x+1) frac{d}{d x}(x^{2}+x+1)-(x^{2}+x+1) frac{d}{d x}(x^{2}-x+1)}{(x^{2}-x+1)^{2}})+frac{2}{sqrt{3}}{frac{(1-x^{2})^{2}}{1+x^{4}-2 x^{2}+3 x^{2}}}{frac{(1-x^{2}) frac{d}{d x}(sqrt{3} x)-sqrt{3} x frac{d}{d x}(1-x^{2})}{(1-x^{2})^{2}}}`

`Rightarrow frac{d y}{d x}=(frac{1}{x^{2}+x+1}){frac{(x^{2}-x+1)(2 x+1)-(x^{2}+x+1)(2 x-1)}{(x^{2}-x+1)}}+frac{2}{sqrt{3}}{frac{(1-x^{2})^{2}}{1+x^{2}+x^{4}}}{frac{(1-x^{2})(sqrt{3})-sqrt{3} x(-2 x)}{(1-x^{2})^{2}}}`

`Rightarrow frac{d y}{d x}=(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^{4}+2 x^{2}+1-x^{2}})+frac{2}{sqrt{3}}(frac{sqrt{3}-sqrt{3} x^{2}+2 sqrt{3} x^{2}}{1+x^{2}+x^{4}})`

`Rightarrow frac{d y}{d x}=(frac{-2 x^{2}+2}{x^{4}+x^{2}+1})+frac{2 sqrt{3}(x^{2}+1)}{sqrt{3}(1+x^{2}+x^{4})}`

`Rightarrow frac{d y}{d x}=frac{2(1-x^{2})}{(x^{4}+x^{2}+1)}+frac{2(x^{2}+1)}{1+x^{2}+x^{4}}`

`Rightarrow frac{d y}{d x}=frac{2(1-x^{2}+x^{2}+1)}{1+x^{2}+x^{4}}`

`Rightarrow frac{d y}{d x}=frac{4}{1+x^{2}+x^{4}}`