Question Number 20871 by j.masanja06@gmail.com last updated on 05/Sep/17
Answered by sma3l2996 last updated on 05/Sep/17
$$\frac{{dy}}{{dx}}=\mathrm{2}\frac{{tan}^{−\mathrm{1}} {x}}{\mathrm{1}+{x}^{\mathrm{2}} }\Leftrightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}=\mathrm{2}{tan}^{−\mathrm{1}} {x} \\ $$$$\frac{{d}}{{dx}}\left(\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}\right)=\mathrm{2}\frac{{d}\left({tan}^{−\mathrm{1}} {x}\right)}{{dx}}=\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$