find dy/dx if xy log(x+y)=1

Answer 1

looks like triple product rule
recall
d(uvw) = uvw' + uwv' + vwu'

I will also assume that you meant ln(x+y)

xy ln(x+y) = 1
xy(1/(x+y)) * (1 + dy/dx) + x ln(x+y) dy/dx + y ln(x+y) = 0

expand the first part to get at the dy/dx
xy(1/(x+y)) + dy/dx xy(1/(x+y)) + dy/dx (x ln(x+y)) = -y ln(x+y) - xy(1/(x+y))
dy/dx [xy(1/(x+y)) + (x ln(x+y)) ] = -y ln(x+y) - xy(1/(x+y))
dy/dx = (-y ln(x+y) - xy(1/(x+y))) / [xy(1/(x+y)) + (x ln(x+y)) ]

leave it like that or simplify as needed.