Question Number 116319 by bemath last updated on 03/Oct/20
$$\mathrm{If}\:\mathrm{z}\:=\:\mathrm{x}^{\mathrm{2}} \:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{y}}{\mathrm{x}}\right),\:\mathrm{find}\:\frac{\partial^{\mathrm{2}} \mathrm{z}}{\partial\mathrm{x}\partial\mathrm{y}}\: \\ $$$$\mathrm{at}\:\left(\mathrm{1},\mathrm{1}\right) \\ $$
Answered by john santu last updated on 03/Oct/20
$$\frac{\partial{z}}{\partial{y}}\:=\:{x}^{\mathrm{2}} .\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }.\frac{\partial}{\partial{y}}\left(\frac{{y}}{{x}}\right)=\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\frac{\partial^{\mathrm{2}} {z}}{\partial{x}\partial{y}}\:=\:\frac{\partial}{\partial{x}}\:\left(\frac{\partial{z}}{\partial{y}}\right)\:=\:\frac{\partial}{\partial{x}}\left(\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−{x}^{\mathrm{3}} \left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mid_{\left(\mathrm{1},\mathrm{1}\right)} \:=\:\frac{\mathrm{1}+\mathrm{3}}{\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\:. \\ $$