How do you find dy/dx by implicit differentiation given x^-2+y^-2=1?

Feb 10, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{3} / {x}^{3}$

Explanation:

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

Using the Power Rule and Implicit Differentiation

$- 2 {x}^{-} 3 - 2 {y}^{-} 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$- 2 {y}^{-} 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{-} 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{-} 3}{- 2 {y}^{-} 3} = - {y}^{3} / {x}^{3}$