Question

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?

Answers

no,

it should be

(3y^2)*dy/dx - (x^2)dy/dx - 2xy = 0 ---->y ' = 1

tangent line y = x + 4

#1

I think the - sign would go

(3y^2)*dy/dx - [ (x^2)dy/dx + 2xy] = 0

#2

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

#3