Question

find y' if 3xy^2+y^3=e^(x+y)

Answer 1

derivative on both sides:
(3xy^2)' = 3y^2 + 6xyy', (y^3)' = 3y^2*y' , (6xy)' = 6y + 6xy', (e^(x+y))' = e^(x+y) * (1 + y')
Then,
3y^2 + 6xyy' + 3y^2y' = e^(x+y)*(1+y')
Then,
\[y' = {(3y^2 - e ^{x+y})} / ( {e^{x+y} - 3y^2 - 6xy})\]