If y = tan - 1 4 x 1 + 5 x 2 + tan - 1 2 + 3 x 3 - 2 x , then d y d x =

$y={\tan }^{-1}\left(\frac{{\displaystyle \frac{4x}{1+5x^2}}+{\displaystyle \frac{2+3x}{3-2x}}}{1-\left({\displaystyle \frac{4x}{1+5x^2}}\right)\left({\displaystyle \frac{2+3x}{3-2x}}\right)}\right)$

$y={\tan }^{-1}\left(\frac{12x-8x^2+2+10x^2+3x+15x^3}{3+15x^2-2x-10x^3-8x-12x^2}\right)$

$={\tan }^{-1}\left(\frac{15x^3+2x^2+15x+2}{-10x^3+3x^2-10x+3}\right)$

$y={\tan }^{-1}\left(\frac{\left(15x+2\right)(x^2+1)}{(3-10x)(x^2+1)}\right)={\tan }^{-1}\left(\frac{15x+2}{3-10x}\right)$

$\frac{dy}{dx}=\frac{1}{1+{\left({\displaystyle \frac{15x+2}{3-10x}}\right)}^2}\times \frac{15(3-10x)+10(15x+2)}{(3-10x{)}^2}$

$=\frac{1}{9+100x^2-60x+225x^2+60x+4}\times \frac{65}{1}$

$=\frac{65}{325x^2+13}=\frac{5}{25x^2+1}$