# How do you differentiate y^2-x^3cos(x) = x^2-y?

Jun 1, 2018

$y ' = \frac{2 x + 3 {x}^{2} \cos \left(x\right) - {x}^{3} \sin \left(x\right)}{2 y + 1}$

#### Explanation:

Using the chain and the product rule
we get

$2 y y ' - 3 {x}^{2} c 9 s \left(x\right) + {x}^{3} \sin \left(x\right) = 2 x - y$
solving for $y '$
$y ' \left(2 y + 1\right) = 2 x + 3 {x}^{2} \cos \left(x\right) - {x}^{3} \sin \left(x\right)$
so
$y ' = \frac{2 x + 3 {x}^{2} \cos \left(x\right) - {x}^{3} \sin \left(x\right)}{2 y + 1}$
for $2 y + 1 \ne 0$