quick question involving natural log?

Question

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

if anyone knows how I would appreciate the help

Answer 1

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Take the derivative of both sides of x^2 + ln(x + y^2) = 3y

you will get

2x + [(1 + 2yy')/(x + y^2)] = 3y'

multiply both sides by (x + y^2)

you will get

2x(x + y^2) + (1 + 2yy') = 3y'(x + y^2)

Solve for y'

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