Question Number 207852 by hardmath last updated on 28/May/24
Answered by MM42 last updated on 28/May/24
$${y}\left({y}+{x}+\mathrm{2}\right)={x}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +{xy}+\mathrm{2}{y}−{x}^{\mathrm{2}} =\mathrm{0}={f}\left({x},{y}\right) \\ $$$${x}=\mathrm{4}\Rightarrow{y}^{\mathrm{2}} +\mathrm{6}{y}−\mathrm{16}=\mathrm{0}\Rightarrow{y}=\mathrm{2} \\ $$$$\frac{{dy}}{{dx}}=−\frac{{f}_{{x}} ^{'} }{{f}'_{{y}} }=\frac{\mathrm{2}{x}−{y}}{\mathrm{2}{y}+{x}+\mathrm{2}} \\ $$$$\overset{{x}=\mathrm{4}\:,{y}=\mathrm{2}} {\Rightarrow}\:\:=\frac{\mathrm{8}−\mathrm{2}}{\mathrm{4}+\mathrm{4}+\mathrm{2}}=\frac{\mathrm{6}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}}\:\checkmark \\ $$$$ \\ $$
Commented by hardmath last updated on 28/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by A5T last updated on 29/May/24
$${x}^{\mathrm{2}} ={y}\left({y}+{x}+\mathrm{2}\right)\Rightarrow{x}^{\mathrm{2}} ={y}^{\mathrm{2}} +{xy}+\mathrm{2}{y} \\ $$$$\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \right)=\frac{{d}}{{dx}}\left({y}^{\mathrm{2}} \right)+\frac{{d}}{{dx}}\left({xy}\right)+\mathrm{2}\frac{{d}}{{dx}}\left({y}\right) \\ $$$$\mathrm{2}{x}=\mathrm{2}{y}\frac{{dy}}{{dx}}+{x}\frac{{dy}}{{dx}}+{y}+\frac{\mathrm{2}{dy}}{{dx}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{2}{x}−{y}}{\mathrm{2}{y}+{x}+\mathrm{2}}\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{2}\left(\mathrm{4}\right)−\mathrm{2}}{\mathrm{2}\left(\mathrm{2}\right)+\mathrm{4}+\mathrm{2}}=\frac{\mathrm{6}}{\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$
Commented by hardmath last updated on 29/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$