If e^x+e^y=(e)^x+y , then dy/dx at(2,2) =?

1 Answer
Jun 18, 2018

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

Explanation:

Given that, #e^x+e^y=e^(x+y)=e^x*e^y#.

Dividing the eqn. by #e^(x+y)#, we get,

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

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

# i.e., e^-y+e^-x=1#.

Diff.ing w.r.t. #x, d/dx(e^-y)+d/dx(e^-x)=0#.

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

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

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

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

N.B. : Since he point #(2,2)# does not lie on the curve

#e^x+e^y=e^(x+y)#, we can not evaluate #[dy/dx]_((2,2)#