# How do you find (dy)/(dx) given (2y-x)^2+3x=0?

Jun 10, 2017

Differentiate each term:

$\frac{d \left({\left(2 y - x\right)}^{2}\right)}{\mathrm{dx}} + \frac{d \left(3 x\right)}{\mathrm{dx}} = \frac{d \left(0\right)}{\mathrm{dx}} \text{ [1]}$

The first term requires the use of the chain rule:

$\frac{d \left(f \left(g \left(x\right)\right)\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \frac{\mathrm{dg}}{\mathrm{dx}} \text{ [2]}$

Let $g = 2 y - x$, then, $f \left(g\right) = {g}^{2}$, $\frac{\mathrm{df}}{\mathrm{dg}} = 2 g$, and $\frac{\mathrm{dg}}{\mathrm{dx}} = 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 1$

Substituting into equation [2]:

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

Reverse the substitution for g:

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

Substitute back into equation [1]:

$2 \left(2 y - x\right) \left(2 \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right) + \frac{d \left(3 x\right)}{\mathrm{dx}} = \frac{d \left(0\right)}{\mathrm{dx}}$

The derivative of the second term is trivial and the derivative of 0 is 0:

$2 \left(2 y - x\right) \left(2 \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right) + 3 = 0$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} - \frac{3}{4 \left(2 y - x\right)}$