Please check my work on this implicit differentiation?
y^3 - yx^2 = 8
Find the slope and tangent line at (-3,1)
(3y^2)*dy/dx - (x^2)dy/dx + 2xy = 0
at x = -3 and y = 1
dy/dx = -1
then to find a tangent line:
(y-1)/(x+3) = -1 and the line is y + x + 2 = 0
Is this correct? If not can you show me where I went wrong and fix it?
it should be
(3y^2)*dy/dx - (x^2)dy/dx - 2xy = 0 ---->y ' = 1
tangent line y = x + 4
I think the - sign would go
(3y^2)*dy/dx - [ (x^2)dy/dx + 2xy] = 0
Correct for the dy/dx term
The line is also correct, although typically the equation is set as something equal to y. So y=-x - 2