Question

How do you find `(dy)/(dx)` given `x^2y^3-3y=x^2`?

Answer 1

`dy/dx = (2x(1-y^3))/(3(x^2y^2-1))`

Explanation:

`x^2y^3 - 3y = x^2`

Differentiating wrt `x` gives;
`d/dx(x^2y^3) - d/dx(3y) = d/dx(x^2)`
`:. d/dx(x^2y^3) - 3dy/dx = 2x`

We can use the product rule:
`d/dx(uv)=u(dv)/dx+v(du)/dx`
for the 1st term to give:

`(x^2)(d/dxy^3) + (y^3)(d/dxx^2) - 3dy/dx = 2x`
`:. (x^2)(d/dxy^3) + (y^3)(2x) - 3dy/dx = 2x`

And we use the chain rule so that we differentiate the 1st term wert `y` rather than `x` to give;

`:. (x^2)(dy/dxd/dyy^3) + 2xy^3 - 3dy/dx = 2x`
`:. (x^2)(dy/dx3y^2) + 2xy^3 - 3dy/dx = 2x`
`:. 3x^2y^2dy/dx + 2xy^3 - 3dy/dx = 2x`
`:. 3x^2y^2dy/dx - 3dy/dx = 2x - 2xy^3`
`:. 3(x^2y^2-1)dy/dx = 2x(1-y^3)`
`:. dy/dx = (2x(1-y^3))/(3(x^2y^2-1))`