Question

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

Answer 1

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

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

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

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

$\frac{dy}{dx}=\frac{1}{1+(5x{)}^2}+0$

$\therefore \frac{dy}{dx}=\frac{1}{1+25x^2}$

Formula:

${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)={\tan }^{-1}x+{\tan }^{-1}y$