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

Dec 23, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{-} 3 - 2 {y}^{-} 2}{1 + 4 x {y}^{-} 3}$

#### Explanation:

$\text{differentiate "2xy^-2" using the "color(blue)"product rule}$

$2 \left(- 2 x {y}^{-} 3. \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{-} 2\right) - 2 {x}^{-} 3 = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow - 4 x {y}^{-} 3 \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {y}^{-} 2 - 2 {x}^{-} 3 = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(1 + 4 x {y}^{-} 3\right) = 2 {x}^{-} 3 - 2 {y}^{-} 2$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{-} 3 - 2 {y}^{-} 2}{1 + 4 x {y}^{-} 3}$