Question

How do you differentiate `xy=cot(xy)`?

Answer 1

`dy/dx = - y/x`

Explanation:

I am assuming that you want to find `dy/dx`.

We shall use both implicit differentiation and chain rule.
`xy = cot(xy)`

Differentiate both sides with respect to `x` to get:
`d/dx (xy) = d/dx (cot(xy))`

Apply chain rule
`d/dx (xy) = -csc^2(xy) d/dx (xy)`
`=> (1 + csc^2(xy)) d/dx (xy) = 0`
`=> d/dx (xy) = 0` (Since `(1 + csc^2(xy))` is always positive)

Apply chain rule again
`y + x dy/dx = 0`
`dy/dx = - y/x`