Question

How do you differentiate `ln(x-y)e^(x-y)=0`?

Answer 1

The only possible solution is:

`y'(x) = 1`

Explanation:

As the exponential function is never null:

`ln(x-y)e^(x-y) = 0 => ln (x-y) = 0 => x-y = 1`

Thus we can make `y` explicit:

`y(x) = (x-1)`

and: `y'(x) = 1`