differentiation
posted by Anamika on .
find dy/dx if xy log(x+y)=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.