Do this please?

Question

Find the slope of the tangent line to the curve
-1x^2+1xy-2y^3=-439
at the point (-1,6)

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Answer 1

Differentiate both sides with respect to x:

d/dx( -x² + xy - 2y³) = d/dx(-439)

-2x + y + (x dy/dx) - (2*3y² dy/dx) = 0

(dy/dx)(x - 6y²) = 2x - y

dy/dx = (2x - y)/(x - 6y²)

Evaluate dy/dx at (-1,6) to get the slope of the tangent line at (-1,6).