Question

How do you implicitly differentiate `xy^2-6x-8y=3`?

Answer 1

`dy/dx = (6-y^2)/(2xy-8)`

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:

`xy^2-6x-8y=3`

Differentiate wrt `x` (applying product rule):

`(x)(d/dxy^2) + (d/dxx)(y^2) - d/dx6x -d/dx8 y = d/dx3`
`:. (x)(2ydy/dx) + (1)(y^2) - 6 -8 dy/dx = 0`
`:. 2xydy/dx + y^2 - 6 -8 dy/dx = 0`
`:. (2xy-8)dy/dx = 6- y^2`
`:. dy/dx = (6- y^2 )/(2xy-8)`

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) = xy^2-6x-8y=3`; Then;

`(partial F)/(partial x) = y^2-6`

`(partial F)/(partial y) = 2xy-8`

And so:

`dy/dx = -(y^2-6)/(2xy-8) = (6-y^2)/(2xy-8)`, as before.