# Solve this

Question:

If $u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$, where $-1 Solution:$u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$and$v=\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$We know,$\frac{\mathrm{du}}{\mathrm{dx}}=\frac{2}{1+\mathrm{x}^{2}}$Using the chain rule of differentiation,$\frac{d v}{d x}=\frac{1}{1+\left(\frac{2 x}{1+x^{2}}\right)^{2}} \cdot \frac{\left(1+x^{2}\right) \cdot(2 x)^{\prime}-\left(1+x^{2}\right)^{\prime} \cdot(2 x)}{\left(1+x^{2}\right)^{2}}=\frac{\left(1+x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}+(2 x)^{2}} \cdot \frac{2\left(1+x^{2}\right)-(2 x)(2 x)}{\left(1+x^{2}\right)^{2}}=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}+(2 x)^{2}}=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}+(2 x)^{2}}$Using Chain Rule of Differentiation,$\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dv}}=\frac{2}{1+x^{2}} \cdot \frac{\left(1+x^{2}\right)^{2}+(2 x)^{2}}{2\left(1-x^{2}\right)}=\frac{\left(1+x^{2}\right)^{2}+(2 x)^{2}}{\left(1+x^{2}\right)\left(1-x^{2}\right)}$Dividing numerator and denominator by$\left(1+x^{2}\right)^{2}$,$\frac{d u}{d v}=\frac{1+\left(\frac{2 x}{1+x^{2}}\right)^{2}}{\frac{1-x^{2}}{1+x^{2}}}=\frac{1+\sin ^{2} u}{\cos u}=\sec u(1+\tan u)\$ (Ans)