Question Number 129031 by oustmuchiya@gmail.com last updated on 12/Jan/21
$${Given}\:{that}\:\boldsymbol{{tan}}^{−\mathrm{1}} \boldsymbol{{x}}\:{show}\:{that}\:\: \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}\:=\:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} } \\ $$
Answered by MJS_new last updated on 12/Jan/21
$${y}=\mathrm{arctan}\:{x} \\ $$$${x}=\mathrm{tan}\:{y} \\ $$$${dx}=\frac{{dy}}{\mathrm{cos}^{\mathrm{2}} \:{y}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{cos}^{\mathrm{2}} \:{y} \\ $$$$\mathrm{cos}\:{y}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{y}}}\overset{{y}=\mathrm{arctan}\:{x}} {=}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$