# How do you implicitly differentiate y/x^4-3/y^2-y=8 ?

Dec 1, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 y {x}^{- 5}}{{x}^{- 4} + 6 {y}^{- 3} - 1}$

#### Explanation:

First, the equation will be rewritten to get $y {x}^{- 4} - 3 {y}^{- 2} - y = 8$

Now we differentiate with respect to $x$:

$\frac{d}{\mathrm{dx}} \left[y {x}^{- 4} - 3 {y}^{- 2} - y\right] = \frac{d}{\mathrm{dx}} \left[8\right]$

$\frac{d}{\mathrm{dx}} \left[y {x}^{- 4}\right] - \frac{d}{\mathrm{dx}} \left[3 {y}^{- 2}\right] - \frac{d}{\mathrm{dx}} \left[y\right] = \frac{d}{\mathrm{dx}} \left[8\right]$

$\frac{d}{\mathrm{dx}} \left[y\right] {x}^{- 4} - 4 y {x}^{- 5} - \frac{d}{\mathrm{dx}} \left[3 {y}^{- 2}\right] - \frac{d}{\mathrm{dx}} \left[y\right] = 0$

Using the chain rule, we get $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dy}} \left[y\right] \cdot \frac{\mathrm{dy}}{\mathrm{dx}} {x}^{- 4} - 4 y {x}^{- 5} - \frac{d}{\mathrm{dy}} \left[3 {y}^{- 2}\right] \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{d}{\mathrm{dy}} \left[y\right] \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} {x}^{- 4} - 4 y {x}^{- 5} + 6 {y}^{- 3} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[{x}^{- 4} + 6 {y}^{- 3} - 1\right] - 4 y {x}^{- 5} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[{x}^{- 4} + 6 {y}^{- 3} - 1\right] = 4 y {x}^{- 5}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 y {x}^{- 5}}{{x}^{- 4} + 6 {y}^{- 3} - 1}$