# How do you implicitly differentiate -3=1/(1-e^y)?

$y ' = 0$

#### Explanation:

transform the equation so that

${e}^{y} = \frac{4}{3}$

differentiate on both sides of the equation

${e}^{y} \cdot y ' = 0$

${e}^{y} y ' = 0$

$y ' = 0$