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

May 19, 2015

This is a bit of a vague question, as it can be differentiated in terms of either $x$ or $y$ or both $x \mathmr{and} y$

I will assume for $x \mathmr{and} y$

${d}^{2} / \left(\mathrm{dx} \mathrm{dy}\right) \cos \left(y\right) - \left({x}^{2} {y}^{3}\right) + 2 y = {d}^{2} / \left(\mathrm{dx} \mathrm{dy}\right) \pi$

we then first differentiate over $y$ and treat $x$ as a constant

$\frac{d}{\mathrm{dx}} - \sin \left(y\right) - \left(3 {x}^{2} {y}^{2}\right) + 2 = \frac{d}{\mathrm{dx}} 0$

then we can differentiate over $x$ and treat $y$ as a constant

$- 1 - \left(6 x {y}^{2}\right) = 0$

$6 x {y}^{2} = - 1$

$x {y}^{2} = - \frac{1}{6}$