# How do you find the derivative of (x^2-y^2)/(3xy)?

Mar 14, 2015

The derivatives are:

$\frac{\partial z}{\partial x} = \frac{{x}^{2} + {y}^{2}}{3 {x}^{2} y}$,

$\frac{\partial z}{\partial y} = - \frac{{x}^{2} + {y}^{2}}{3 x {y}^{2}}$.

This is because:

$\frac{\partial z}{\partial x} = \frac{1}{3} \cdot \frac{2 x \cdot x y - \left({x}^{2} - {y}^{2}\right) \cdot y}{{x}^{2} {y}^{2}} = \frac{2 {x}^{2} y - {x}^{2} y + {y}^{3}}{3 {x}^{2} {y}^{2}} =$

$= \frac{{x}^{2} y + {y}^{3}}{3 {x}^{2} {y}^{2}} = \frac{{x}^{2} + {y}^{2}}{3 {x}^{2} y}$;

and

$\frac{\partial z}{\partial y} = \frac{1}{3} \cdot \frac{- 2 y \cdot x y - \left({x}^{2} - {y}^{2}\right) x}{{x}^{2} {y}^{2}} = \frac{- 2 x {y}^{2} - {x}^{3} + x {y}^{2}}{3 {x}^{2} {y}^{2}} =$

$= - \frac{{x}^{3} + x {y}^{2}}{3 {x}^{2} {y}^{2}} = - \frac{{x}^{2} + {y}^{2}}{3 x {y}^{2}}$.