How do you differentiate #e^x+e^y=e^(x+y)#?

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Answer 1 · 2016-06-08T09:59:30

#(dy)/(dx)=(e^x(e^y-1))/(e^y(1-e^x))#

Differentiating #e^x+e^y=e^(x+y)#

#e^x+e^y(dy)/(dx)=e^(x+y)(1+(dy)/(dx))#

or #e^x+e^y(dy)/(dx)=e^(x+y)+e^(x+y)(dy)/(dx)#

or #e^y(dy)/(dx)-e^(x+y)(dy)/(dx)=e^(x+y)-e^x#

or #(e^y-e^(x+y))(dy)/(dx)=(e^(x+y)-e^x)#

or #(dy)/(dx)=(e^(x+y)-e^x)/(e^y-e^(x+y))=(e^x(e^y-1))/(e^y(1-e^x))#

Answer 2 · 2018-07-13T19:28:56

# dy/dx=-e^(y-x)#.

# e^x+e^y=e^(x+y)=e^x*e^y#.

#:. (e^x+e^y)/(e^x*e^y)=1#.

#:. e^x/(e^x*e^y)+e^y/(e^x*e^y)=1#.

#:. e^-y+e^-x=1#.

Diff.ing w.r.t. #x#, we get,

# d/dx(e^-y)+d/dx(e^-x)=d/dx(1)#.

#:. d/dy(e^-y)*d/dx(-y)+e^-x*d/dx(-x)=0......[because," the Chain Rule]"#.

#:. e^-y(-dy/dx)+e^-x*-1=0#.

#:. -dy/dx=e^-x/e^-y=e^(y-x)#.

# rArr dy/dx=-e^(y-x)#, as Respected Shwetank Mauria has

readily derived!