Question Number 62291 by rajesh4661kumar@gamil.com last updated on 19/Jun/19
Commented by maxmathsup by imad last updated on 19/Jun/19
$$\Rightarrow\left({e}^{{x}} −\mathrm{1}\right){e}^{{y}} \:={e}^{{x}} \:\Rightarrow{e}^{{y}} \:=\frac{{e}^{{x}} }{{e}^{{x}} −\mathrm{1}}\:\Rightarrow{y}\:={ln}\left(\frac{{e}^{{x}} }{{e}^{{x}} −\mathrm{1}}\right)\:\Rightarrow{y}\:={x}−{ln}\left({e}^{{x}} −\mathrm{1}\right)\:\Rightarrow \\ $$$$\frac{{dy}}{{dx}}\:=\mathrm{1}−\frac{{e}^{{x}} }{{e}^{{x}} −\mathrm{1}}\:=\frac{{e}^{{x}} −\mathrm{1}−{e}^{{x}} }{{e}^{{x}} −\mathrm{1}}\:=\frac{−\mathrm{1}}{{e}^{{x}} −\mathrm{1}}\:. \\ $$
Answered by tanmay last updated on 19/Jun/19
$${e}^{{x}} +{e}^{{y}} ×\frac{{dy}}{{dx}}={e}^{{x}+{y}} ×\left(\mathrm{1}+\frac{{dy}}{{dx}}\right) \\ $$$${e}^{{x}} −{e}^{{x}+{y}} ={e}^{{x}+{y}} ×\frac{{dy}}{{dx}}−{e}^{{y}} ×\frac{{dy}}{{dx}} \\ $$$${e}^{{x}} \left(\mathrm{1}−{e}^{{y}} \right)=\frac{{dy}}{{dx}}×{e}^{{y}} ×\left({e}^{{x}} −\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{{e}^{{x}} \left(\mathrm{1}−{e}^{{y}} \right)}{{e}^{{y}} \left({e}^{{x}} −\mathrm{1}\right)}=\frac{{e}^{{x}} −{e}^{{x}+{y}} }{{e}^{{x}+{y}} −{e}^{{y}} }=\frac{−{e}^{{y}} }{{e}^{{x}} }=−{e}^{{y}−{x}} \\ $$
Commented by rajesh4661kumar@gamil.com last updated on 19/Jun/19
$${thans}\:{ji} \\ $$
Commented by tanmay last updated on 19/Jun/19
$${most}\:{welcome}\:{sir}... \\ $$