Find the slope of the curve at the given point y(3xy-x^2)=16 at (2,2)

Question

Please help me answer my h.w, thank you.

Answer 1

Problems such as this, where we have f(x,y) = 0, require implicit differentiation.

Just so we don't get lost, let's just expand the left side.

3xy^2 - x^2y = 16

(3y^2 -2xy)dx + (6xy - x^2) dy = 0

(6xy - x^2) dy = - (3y^2 -2xy)dx

dy/dx = - (3y^2 -2xy)/(6xy - x^2) or

dy/dx = (2xy-3y^2)/(6xy - x^2)

Now, substitute x = 2 and y = 2

dy/dx = (2*2^2 -3*2^2)/(6*2^2-2^2) = -1/5

Answer 2

Use implicit differentiation to get the derivative dy/dx= (2xy-3y²)/(6xy-x²)

At (2,2) the derivative is -1/5.

You will need to remember the rule for the derivative of a product in order to get dy/dx.