If $y= an ^{-1}\left(\frac{2+3 x}{3-2 x} ight)+ an ^{-1}\left(\frac{4 x}{1+5 x^2} ight)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

Question

If $y=	an ^{-1}\left(\frac{2+3 x}{3-2 x}ight)+	an ^{-1}\left(\frac{4 x}{1+5 x^2}ight)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

Aicharya · Accepted Answer

$\begin{aligned} & y=	an ^{-1}\left(\frac{2+3 x}{3-2 x}ight)+	an ^{-1}\left(\frac{4 x}{1+5 x^2}ight) \ & =	an ^{-1}\left(\frac{\frac{2}{3}+x}{1-\frac{2}{3} x}ight)+	an ^{-1}\left(\frac{5 x-x}{1+5 x^2}ight) \ & =	an ^{-1} \frac{2}{3}+	an ^{-1} x+	an ^{-1} 5 x-	an ^{-1} x \ 	herefore \quad & y=	an ^{-1} \frac{2}{3}+	an ^{-1} 5 x \ 	herefore \quad & \frac{d y}{d x}=\frac{1}{1+(5 x)^2} \cdot 5=\frac{5}{1+25 x^2}\end{aligned}$

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

Solution