Question

How do you find `(dy)/(dx)` given `4x^2+4xy=-5x^3y+4`?

Answer 1

`dy/dx = -(15x^2y + 8x + 4y)/(4x + 5x^3)`

Explanation:

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

However, we only differentiate explicit functions of `y` wrt `x`. But if we apply the chain rule we can differentiate an implicit function of `y` wrt `y` but we must also multiply the result by `dy/dx`.

Example:

`d/dx(y^2) = d/dy(y^2)dy/dx = 2ydy/dx`

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

Now, we have:

`4x^2+4xy = -5x^3y+4`

Implicitly differentiating wrt `x` (applying product rule):

`8x + (4x)(dy/dx) + (4)(y) = (-5x^3)(dy/dx) + (-15x^2)(y)`

`:. 8x + 4xdy/dx + 4y = -5x^3dy/dx -15x^2y`

`:. 4xdy/dx + 5x^3dy/dx = -15x^2y -8x -4y`

`:. (4x + 5x^3)dy/dx = -(15x^2y + 8x + 4y)`

`:. dy/dx = -(15x^2y + 8x + 4y)/(4x + 5x^3)`

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^2+4xy + 5x^3y-4`; Then;

`(partial F)/(partial x) = 8x +4y + 15x^2y`

`(partial F)/(partial y) = 4x + 5x^3`

And so:

`dy/dx = -(8x +4y + 15x^2y)/(4x + 5x^3)`, as before.