Mathematics | VITEEE 2019 | Discussion Page

If y = $\tan^{-1}\frac{4x}{1+5x^{2}}+\tan^{-1}\frac{2+3x}{3-2x}$ then $\frac{dy}{dx}$ =

$\frac{1}{1+25x^{2}}+\frac{2}{1+x^{2}}$

$\frac{5}{1+25x^{2}}+\frac{2}{1+x^{2}}$

$\frac{5}{1+25x^{2}}$

$\frac{1}{1+25x^{2}}$

Option C

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

= $\tan^{-1}\frac{5x-x}{1+5x.x}+\tan^{-1}\frac{\frac{2}{3}+x}{1-\frac{2}{3}.x}$

= $\tan^{-1}5x-\tan^{-1}x+\tan^{-1}\frac{2}{3}+\tan^{-1}x$

then $\frac{dy}{dx}$ = $\frac{5}{1+25x^{2}}$

Discussion Forum