Question

Find dy/dx by implicit differentiation.
2x^3 + x^2y - xy^3 = 1

Answer 1

d/dx= Dx; dy/dx = y'; and dx/dx=x'=1

Dx(2x^3) = 6x^2(x')
Dx(yx^2) = 2xy(x') + y'x^2
Dx(xy^3)= y^3(x') + x 3y^2 y'
Dx(1)= 0(x')

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

y'=(-6x^2 -2xy +y^3)/(x^2 -3xy^2)

If im lucky, that works out :)

Answer 2

thank u!