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

Jan 12, 2017

The only possible solution is:

$y ' \left(x\right) = 1$

#### Explanation:

As the exponential function is never null:

$\ln \left(x - y\right) {e}^{x - y} = 0 \implies \ln \left(x - y\right) = 0 \implies x - y = 1$

Thus we can make $y$ explicit:

$y \left(x\right) = \left(x - 1\right)$

and: $y ' \left(x\right) = 1$