Determine the slope of the graph of the relation?

Question

2x^2-3xy+y^3= -1 at the point (2,-3)

Answer 1

Take derivative according to x

4x-(3y+y'.3x)+3y^2.y'=0
Plug in x and y value of the point

4.2-(3(-3)+y'.3.2)+3(-3)^2 . y'=0

8-(-9+6y')+27y'=0
17+21y'=0
21y'=-17

The slope at the given point is
y'=-17/21

Answer 2

d/dx(2x^2-3xy+y^3)=0

solving the above eqn gives us the eqn :


dy/dx = ( 3y - 4x )/ ( 3x + 3y^2 )

then dy/dx at (2,-3) = 18/33