Question

How do you implicitly differentiate `-1=xycot^2(x-y)`?

Answer 1

`dy/dx = {2xycot(x-y)csc^2(x-y) - ycot^2(x-y)}/{2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}`

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:

`-1=xycot^2(x-y)`

Differentiate wrt `x` (applying triple product rule):

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

`:. 0 = (x)(y)(2cot(x-y)(-csc^2(x-y))(1-dy/dx)) + (x)(dy/dx)(cot^2(x-y)) + (1)(y)(cot^2(x-y))`

`:. 0 = -2xycot(x-y)csc^2(x-y)(1-dy/dx) + xdy/dxcot^2(x-y) + ycot^2(x-y)`

`:. 2xycot(x-y)csc^2(x-y)dy/dx + xdy/dx(cot^2(x-y) = 2xycot(x-y)csc^2(x-y) - ycot^2(x-y)`

`:. {2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}dy/dx = 2xycot(x-y)csc^2(x-y) - ycot^2(x-y)`

`:. dy/dx = {2xycot(x-y)csc^2(x-y) - ycot^2(x-y)}/{2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}`

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) =xycot^2(x-y)+1`; Then using the product rule;

`(partial F)/(partial x) = (xy)((partial)/(partial x) cot^2(x-y)) + ((partial)/(partial x) xy)(cot^2(x-y))`
`\ \ \ \ \ \ \ = (xy)(2cot(x-y)(-csc^2(x-y))) + (y)(cot^2(x-y))`
`\ \ \ \ \ \ \ = -2xycot(x-y)csc^2(x-y) + ycot^2(x-y)`

`(partial F)/(partial x) = (xy)((partial)/(partial y) cot^2(x-y)) + ((partial)/(partial y) xy)(cot^2(x-y))`
`\ \ \ \ \ \ \ = (xy)(2cot(x-y)(-csc^2(x-y)(-1)) + (x)(cot^2(x-y))`
`\ \ \ \ \ \ \ = 2xycot(x-y)csc^2(x-y) + xcot^2(x-y)`

And so:

`dy/dx = -{-2xycot(x-y)csc^2(x-y) + ycot^2(x-y)}/{2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}`

`\ \ \ \ \ = {2xycot(x-y)csc^2(x-y) - ycot^2(x-y)}/{2xycot(x-y)csc^2(x-y) + xcot^2(x-y)}`

Here we get an immediate implicit function for the derivative, so here if you are familiar with partial differentiation the solution is much easier