Question

X^3y-2x^2+y^4=8
find dy/dx

Answer 1

`3x^2y +x^3(dy)/(dx) -4x + 4y(dy)/(dx) = 0\\(dy)/(dx) (x^3 +4y) = 4x -3x^2y\\(dy)/(dx) = (4x-3x^2y)/(x^3+4y)`

Answer 2

x^3y-2x^2+y^4=8

take the derivative of each part, using the product rule on the first term

x^3 dy +y*3x^2dx -4xdx+4y^3 dy = 0

gather the dy and dx terms together

dy (x^3+4y^3) -dx(-3x^2y+4x) =0

dy terms on the left, dx terms on the right

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

dy divided by dx

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

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