# What is the implicit derivative of 1= xy-e^(xy) ?

##### 1 Answer
Jun 21, 2016

$y ' = - \frac{y}{x} , \setminus q \quad x , y \setminus \ne 0$

#### Explanation:

$0 = \frac{d}{\mathrm{dx}} \left(x y\right) - \frac{d}{\mathrm{dx}} {e}^{x y}$

first term, using product rule:

$\frac{d}{\mathrm{dx}} \left(x y\right) = y + x y '$

second term using chain rule:

$\frac{d}{\mathrm{dx}} {e}^{x y} = \frac{d}{\mathrm{dx}} \left(x y\right) \cdot {e}^{x y}$

with product rule this is $\frac{d}{\mathrm{dx}} \left(x y\right) \cdot {e}^{x y} = \left(y + x y '\right) \cdot {e}^{x y}$

combining terms

$\frac{d}{\mathrm{dx}} \left(x y\right) - \frac{d}{\mathrm{dx}} {e}^{x y} = \left(y + x y '\right) - \left(y + x y '\right) \cdot {e}^{x y}$

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

ignoring trivial solution $\left(1 - {e}^{x y}\right) = 0 \setminus \implies x = 0 , \mathmr{and} y = 0$ leaves

$y + x y ' = 0 \setminus \implies y ' = - \frac{y}{x} , \setminus q \quad x , y \setminus \ne 0$