# How do you implicitly differentiate x-y^2=ysin(x)+y?

Mar 21, 2016

You can do it like this:

#### Explanation:

$x - {y}^{2} = y \sin \left(x\right) + y$

Differentiate both sides implicitly with respect to $x$:

I will use the power rule, the chain rule and the product rule:

$D \left(x - {y}^{2}\right) = D \left[y \sin \left(x\right) + y\right]$

$\therefore 1 - 2 y . y ' = y \cos \left(x\right) + \sin \left(x\right) . y ' + y '$

$\therefore 1 - y \cos \left(x\right) = 2 y . y ' + \sin \left(x\right) . y ' + y '$

$\therefore 1 - y \cos \left(x\right) = y ' \left(2 y + \sin \left(x\right) + 1\right)$

$\therefore y ' = \frac{\left(1 - y \cos \left(x\right)\right)}{\left(2 y + \sin \left(x\right) + 1\right)}$