# What is 2xy differentiated implicitly?

Sep 2, 2015

$y ' = \frac{2 y}{1 - 2 x}$

#### Explanation:

The question does not specify with respect to what so I'll assume y is a function of x.

Use the product rule:

$y ' = d \frac{\left(u . v\right)}{\mathrm{dx}} = v . \frac{\mathrm{du}}{\mathrm{dx}} + u . \frac{\mathrm{dv}}{\mathrm{dx}}$

So:

$y ' = 2 x . y ' + y 2. \frac{\mathrm{dx}}{\mathrm{dx}}$

$y ' = 2 x . y ' + 2 y$

$y ' = \frac{2 y}{1 - 2 x}$

Jul 20, 2018

The answer is $= - \frac{y}{x}$

#### Explanation:

The function is

$f \left(x , y\right) = 2 x y$

The partial derivatives are

$\frac{\partial f}{\partial x} = 2 y$

$\frac{\partial f}{\partial y} = 2 x$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = - \frac{2 y}{2 x} = - \frac{y}{x}$