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Question-62291




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

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