# How do you differentiate x^7 - 8xy + y^4 = 7?

Apr 21, 2018

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{8 y - 7 {x}^{6}}{4 {y}^{3} - 8 x}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({x}^{7} - 8 x y + {y}^{4}\right) = \frac{d}{\mathrm{dx}} \left(7\right)$

$7 {x}^{6} - 8 y - 8 x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$4 {y}^{3} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 8 x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 8 y - 7 {x}^{6}$

$\left(4 {y}^{3} - 8 x\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 8 y - 7 {x}^{6}$

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{8 y - 7 {x}^{6}}{4 {y}^{3} - 8 x}$