Problem solving process:

Deduction on both sides of the equation:

y+xy'=e^(x+y)( 1+y')

y+xy'=e^(x+y)+y'e^(x+y)

y'[x-e^(x+y)]=e^(x+y)-y

The final result is: y' = [e (x+y)-y]/[x-e (x+y)]

If the equation F(x, y)=0 can determine that y is a function of x, then the function expressed in this way is called an implicit function. The function means that in a certain change process, two variables X and Y, for each value of X in a certain range, Y has a certain value corresponding to it, Y is a function of X, and the relationship is expressed as y=f(x), that is, an explicit function.

Extended data:

If the function is not limited to continuity, the symbol in the formula can change with x, so there are infinite solutions; If the continuity is limited, there are only two solutions (one is always positive and the other is always negative); If it is limited to differentiability, X = 1 should be excluded, so the domain of the function should be an open interval (-1

What additional conditions can make the original equation determine a unique function y= on the premise that the function F(x, y) is continuously differentiable near the point suitable for the original equation? (x), which is not only single-valued continuous, but also continuously differentiable's, and its derivative is completely determined by. It is not only necessary but also sufficient to use the existence theorem of implicit function to determine such a condition.

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