If y = tan - 1 a x - b b x + a , then y ' =

$y={\tan }^{-1}\left\{\frac{ax-b}{bx+a}\right\}$

$\frac{dy}{dx}=\frac{1}{1+{\left\{\frac{ax-b}{bx+a}\right\}}^2}\times \left(\frac{\left(bx+a\right)a-\left(ax-b\right)b}{{\left(bx+a\right)}^2}\right)$

$\frac{dy}{dx}=\frac{1}{1+{\left\{\frac{ax-b}{bx+a}\right\}}^2}\times \frac{a^2+b^2}{{\left(bx+a\right)}^2}$

$\frac{dy}{dx}=\frac{a^2+b^2}{b^2{x}^2+a^2+2abx+a^2{x}^2+b^2-2abx}=\frac{a^2+b^2}{x^2\left(a^2+b^2\right)+\left(a^2+b^2\right)}$

$\frac{dy}{dx}=\frac{1}{x^2+1}$