If $y= an ^{-1}\left[\frac{1}{1+x+x^2} ight]+ an ^{-1}\left[\frac{1}{x^2+3 x+3} ight], x0$, then $\frac{d y}{d x}=$

Question

If $y=	an ^{-1}\left[\frac{1}{1+x+x^2}ight]+	an ^{-1}\left[\frac{1}{x^2+3 x+3}ight], x>0$, then $\frac{d y}{d x}=$

Aicharya · Accepted Answer

$\begin{aligned} & y=	an ^{-1}\left[\frac{1}{1+x+x^2}ight]+	an ^{-1}\left[\frac{1}{x^2+3 x+3}ight] \ & =	an ^{-1}\left[\frac{1}{1+x(1+x)}ight]+	an ^{-1}\left[\frac{1}{1+(x+2)(x+1)}ight] \ & =	an ^{-1}\left[\frac{(x+1)-1}{1+(x+1) x)}ight]+	an ^{-1}\left[\frac{(x+2)-(x+1)}{1+(x+2)(x+1)}ight] \ & =	an ^{-1}(\mathrm{x}+1)-	an ^{-1}(\mathrm{x}+2)-	an ^{-1}(\mathrm{x}+1) \ & =	an ^{-1}(\mathrm{x}+2)-	an ^{-1}(\mathrm{x}) \ & 	herefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+(\mathrm{x}+2)^2}-\frac{1}{1+\mathrm{x}^2} \ & \end{aligned}$

If $y=\tan ^{-1}\left[\frac{1}{1+x+x^2}\right]+\tan ^{-1}\left[\frac{1}{x^2+3 x+3}\right], x>0$, then…

Solution