Question

How do you find `(dy)/(dx)` given `-2xy^2-3x^2y^3+3=4x^3`?

Answer 1

`dy/dx = -(12x^2 + 2y^2 +6xy^3)/(4xy+9x^2y^2)`

Explanation:

When we differentiate `y` wrt `x` we get `dy/dx`.

However, we cannot differentiate a non implicit function of `y` wrt `x`. But if we apply the chain rule we can differentiate a function of `y` wrt `y` but we must also multiply the result by `dy/dx`.

When this is done in situ it is known as implicit differentiation.

We have:

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

Differentiate wrt `x` (applying product rule):

`(-2x)(2ydy/dx) + (-2)(y^2) + (- 3x^2)(3y^2dy/dx) + (- 6x)(y^3) = 12x^2`

`:. -4xydy/dx -2y^2 - 9x^2y^2dy/dx - 6xy^3 = 12x^2`

`:. -(4xy+9x^2y^2)dy/dx = 12x^2 + 6xy^3 + 2y^2`

`:. dy/dx = -(12x^2 + 6xy^3 + 2y^2)/(4xy+9x^2y^2)`

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

`(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))`

So Let `F(x,y) = 4x^3 + 2xy^2 + 3x^2y^3 - 3`; Then;

`(partial F)/(partial x) = 12x^2 + 2y^2 +6xy^3`

`(partial F)/(partial y) = 4xy+9x^2y^2`

And so:

`dy/dx = -(12x^2 + 2y^2 +6xy^3)/(4xy+9x^2y^2)`, as before.