Question

How do you find `dy/dx` by implicit differentiation of `x^3-xy+y^2=4`?

Answer 1

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

Explanation:

`x^3-xy+y^2=4`

We differentiate everything wrt `x`:

`d/dx(x^3) - d/dx(xy) + d/dx(y^2) = d/dx(4)`

We can just deal with the `x^3` and the constant term;

`3x^2 - d/dx(xy) + d/dx(y^2) = 0`

For the second term we apply the product rule;

`3x^2 - { (x)(d/dx(y)) +(d/dx(x))(y } + d/dx(y^2) = 0`
`:. 3x^2 - { xdy/dx +(1)(y) } + d/dx(y^2) = 0`
`:. 3x^2 - xdy/dx + y + d/dx(y^2) = 0`

And for the last term we use the chain rule so that we can differentiate wrt `y` (this is the "Implicit" part of the differentiation

`3x^2 - xdy/dx + y + dy/dxd/dy(y^2) = 0`
`:. 3x^2 - xdy/dx + y + dy/dx2y = 0`
`:. xdy/dx - 2ydy/dx= 3x^2 + y`
`:. (x-3y)dy/dx = 3x^2 + y`
`:. dy/dx = (3x^2 + y)/(x-3y)`