Question

How do you find `dy/dx` by implicit differentiation of `y^2=(x^2-4)/(x^2+4)` and evaluate at point (2,0)?

Answer 1

`dy/dx` does not exist at `y=0`. Finding #dy/dx takes a bit more work.

Explanation:

Beginning with `y^2=(x^2-4)/(x^2+4)`

we differentiate to get

`2y dy/dx = "some function of "x`.

But then `dy/dx` requires dividing by `y`, so it is not defined at `y = 0`

In detail:

`2ydy/dx = (2x(x^2+4)-2x(x^2-4))/(x^2+4)^2`

`2ydy/dx = (16x)/(x^2+4)^2`

`dy/dx = (8x)/(y(x^2+4)^2)`

Here is the graph of the equation.

graph{y^2-(x^2-4)/(x^2+4)=0 [-8.077, 7.724, -4.394, 3.51]}