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