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))

faster, im not sure why but i know it is faster