Use implicit differentiation to find the slope of the tangent line to the curve defined by xy^3+2xy=60 at the point (5,2)

Question

I would appreciate if all work is shown with detailed steps!

Provide the answer to the slope of the tangent line to the curve at the given point is _____.

Answer 1

if the function is xy³ + 2xy = 60

Now differentiate y³ + 2xy²y' + 2y + 2xy' = 0

Then y'(2x + 2xy²) = -y³ - 2y or y' = dy/dx = (-y³ - 2y)/(2x + 2xy²)

Substituting

dy/dx @ (5,2) = ( -2³ - 2*2)/(2*5 + 2*5*(2)² ) = -12/50 = -6/25 = slope of tangent line