Question

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

Answer 1

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

`y+y^3=x^2`

Differentiate wrt `x`:

`\ \ \ \ dy/dx+3y^2dy/dx = 2x`

`:. dy/dx(1+3y^2) = 2x`

`:. dy/dx = (2x)/(1+3y^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) = y+y^3-x^2`; Then;

`(partial F)/(partial x)=-2x \ \ \` and `\ \ \ (partial F)/(partial y) = 1+3y^2`

And so:

`dy/dx = -(-2x) / (1+3y^2) = (2x) / (1+3y^2)`, as before.