Calculus Credit problem?

Question

Given the following expression, find dy/dx implicity:

ln(x+y)=xy^2


Any help would be great

Answer 1

for d/dx ln(x+y)=(dy/dx)*(1/(x+y))

for d/dx (xy^2)=y^2+x(2y)(dy/dx)

so (dy/dx)*(1/(x+y))=y^2+x(2y)(dy/dx)
.

Answer 2

1/(x+y) *(1+y')=y^2 +2xyy'
1+y'= (y^2+2xyy')( x+y)= xy^2 +2x^2yy'+y^3+2xy^2y'
y'(1-2xy^2-2x^2y) = xy^2 +y^3-1

Answer 3

[1/(x+y)][1+dy/dx]=2xy(dy/dx) + y^2
1+dy/dx=2x^2y(dy/dx)+ 2xy^2(dy/dx)+xy^2+y^3
dy/dx(1-2x^2y-2xy^2)
=xy^2+y^3-1
dy/dx=
[xy^2+y^3-1]/[1-2x^2y-2xy^2]

Answer 4

I wish I could have had this question/answer thing when I was in college

Answer 5

ln(x+y)=xy^2
1+dy/dx/(x+y)=y^2+2xydy/dx
1+dy/dx=(x+y)(y^2+2xydy/dx)
dy/dx(1-2xy(x+y))=(x+y)(y^2)-1
dy/dx=
[(x+y)(y^2)-1]/(1-2xy(x+y))