Question

How do you implicitly differentiate `ln(xy)=x-y`?

Answer 1

See the explanation section, below.

Explanation:

`d/dx(ln(xy)) = d/dx(x-y)`

`1/(xy) d/dx(xy) = 1-dy/dx`

`1/(xy)(y+xdy/dx) = 1-dy/dx`

`y+xdy/dx = xy - xydy/dx`

`(x+xy) dy/dx = xy-y`

`dy/dx = (xy-y)/(x+xy)`