Find dy/dx by implicit differentiation. Square root xy = 8 + x^2y

((4xysqrt (xy)) - y) / (x-2x^2sqrt (xy))

Step-by-step explanation:

Treat y as a function of x and use the chain rule.

Use the chain rule and product rule:

(y+xy') / 2sqrt (xy) = d/dx[8+x^2y]

(y+xy') / 2sqrt (xy) = d/dx[x^2y]

Use product rule

(y+xy') / 2sqrt (xy) = 2xy + x^2y'

Now solve for y'

y+xy' = (2xy) (2sqrt (xy)) + (x^2y') (2sqrt (xy))

xy'-2x^2y'sqrt (xy) = (2xy) (2sqrt (xy)) - y

y' (x-2x^2sqrt (xy))

y' = ((4xysqrt (xy)) - y) / (x-2x^2sqrt (xy))