Question

How do you find `dy/dx` by implicit differentiation of `x^2-y^2=16`?

Answer 1

`dy/dx = x/y`

Explanation:

If an equation does not express `y` explicitly in terms of `x` as in `y=f(x)`, and instead we have `g(y)=f(x)`, then when we differentiate we apply the chain rule so that we differentiate `g(y)` wrt `y` rather than `x` as in:

`g(y) = f(x)`
`:. d/dx g(y) = d/dx f(x)`
`:. dy/dx d/dy g(y) = f'(x)`
`:. g'(y) dy/dx= f'(x)`

This will typically result in `dy/dx` being a function of `x` and `y`, rather than `dy/dx` being a function of `x` alone as we usually get

So for `x^2 - y^2 = 16` we have:

`d/dx(x^2) - d/dx(y^2) = d/dx(16)`
`:. 2x - dy/dxd/dy(y^2) = 0`
`:. 2x - 2ydy/dx = 0`
`:. 2ydy/dx = 2x`
`:. dy/dx = x/y`