# Find dy/dx if y^2=x(x-1)^2?

May 25, 2017

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

#### Explanation:

We have:

${y}^{2} = x {\left(x - 1\right)}^{2}$

Differentiating LHS implicitly, and RHS using the product rule along with the chain rule we get:

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

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

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

$\therefore 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(x - 1\right) \left(2 x + \left(x - 1\right)\right)$

$\therefore 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(x - 1\right) \left(3 x - 1\right)$

$\therefore \setminus \setminus \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(x - 1\right) \left(3 x - 1\right)}{2 y}$