Question

Need help with implicit differentiation: Find dy/dx for x+ ln(y) =x^2y^3

Answer 1

(x)'=1
(ln(y))'=y'/y
(x^2y^3)'=(x^2)'y^3+x^2(y^3)'

Answer 2

k I'm at that part- i guess im having trouble getting y' by itsself lol- i need to go back to algebra

Answer 3

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

so dy/dx = (1 - 2xy^3)/ (1/y - 3x^2y^2)

Answer 4

ah ok i see it now- thank you