# How do you find dy/dx by implicit differentiation given x^5=y^2-y+1?

May 21, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4}}{2 y - 1}$

#### Explanation:

Differentiate everything wrt$\text{ "x" }$ then rearrange for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \left({x}^{5} = {y}^{2} - y + 1\right)$

$\frac{d}{\mathrm{dx}} \left({x}^{5}\right) = \frac{d}{\mathrm{dx}} \left({y}^{2}\right) - \frac{d}{\mathrm{dx}} \left(y\right) + \frac{d}{\mathrm{dx}} \left(1\right)$

$5 {x}^{4} = 2 y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} + 0$

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[2 y - 1\right] = 5 {x}^{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4}}{2 y - 1}$