# Differentiate x^4+y^4=8xy?

Sep 15, 2017

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

#### Explanation:

${x}^{4} + {y}^{4} = 8 x y$

$4 {x}^{3} + 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 8 \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot 1\right)$ [Power rule and Product rule]

$4 \left({x}^{3} + {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 8 \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right)$

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

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

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

Sep 15, 2017

$\frac{{x}^{3} - 2 y}{2 x - {y}^{3}}$

#### Explanation:

Using the chain rule on the left side and product rule on the right... we get...
$4 {x}^{3} \mathrm{dx} + 4 {y}^{3} \mathrm{dy} = 8 \left(\mathrm{dx} \cdot y + \mathrm{dy} \cdot x\right)$
Divide both sides by 4
${x}^{3} \mathrm{dx} + {y}^{3} \mathrm{dy} = 2 y \mathrm{dx} + 2 x \mathrm{dy}$
Divide both sides by $\mathrm{dx}$
${x}^{3} + {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 y + 2 x \frac{\mathrm{dy}}{\mathrm{dx}}$
Isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$s and non-$\frac{\mathrm{dy}}{\mathrm{dx}}$s
$\frac{\mathrm{dy}}{\mathrm{dx}} \left({y}^{3} - 2 x\right) = 2 y - {x}^{3}$
Divide both sides by ${y}^{3} - 2 x$ to isolate just the $\frac{\mathrm{dy}}{\mathrm{dx}}$.
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 y - {x}^{3}}{{y}^{3} - 2 x}$
Multiply numerator and denominator by $- 1$.
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{3} - 2 y}{2 x - {y}^{3}}$
And you're done!