Question Number 150839 by mathdanisur last updated on 15/Aug/21
$$\mathrm{e}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{for}\:\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$
Answered by peter frank last updated on 15/Aug/21
$${e}^{{x}} +\frac{{dy}}{{dx}}={x}^{\mathrm{2}} \mathrm{2}{y}\frac{{dy}}{{dx}}+{y}^{\mathrm{2}} \mathrm{2}{x} \\ $$$${e}^{{x}} −\mathrm{2}{xy}^{\mathrm{2}} =\frac{{dy}}{{dx}}\left(\mathrm{2}{x}^{\mathrm{2}} {y}−\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{e}^{{x}} −\mathrm{2}{xy}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} {y}−\mathrm{1}} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{ser} \\ $$
Answered by amin96 last updated on 15/Aug/21
$$\frac{{d}\left({e}^{{x}} \right)}{{dx}}+\frac{{dy}}{{dx}}={x}^{\mathrm{2}} \frac{{d}\left({y}^{\mathrm{2}} \right)}{{dx}}+{y}^{\mathrm{2}} \frac{{d}\left({x}^{\mathrm{2}} \right)}{{dx}} \\ $$$${e}^{{x}} +\frac{{dy}}{{dx}}=\mathrm{2}{x}^{\mathrm{2}} {y}\frac{{dy}}{{dx}}+\mathrm{2}{xy}^{\mathrm{2}} \:\:\Rightarrow\:\:\left\{\:{y}={f}\left({x}\right)\:\:\frac{{d}\left({y}^{{n}} \right)}{{dx}}={ny}^{{n}−\mathrm{1}} \frac{{dy}}{{dx}}\right\} \\ $$$${e}^{{x}} +\frac{{dy}}{{dx}}\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} {y}\right)=\mathrm{2}{xy}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{xy}^{\mathrm{2}} −{e}^{{x}} }{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} {y}} \\ $$
Commented by mathdanisur last updated on 15/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$