# If (x-y)^3 = A(x + y), how do you prove that (2x + y)dy/dx = x + 2y?

May 5, 2015

If ${\left(x - y\right)}^{3} = A \left(x + y\right)$, prove that $\left(2 x + y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = x + 2 y$

Before differentiating, let's isolate the constant $A$

${\left(x - y\right)}^{3} / \left(x + y\right) = A$

Now, use the quotient rule to differentiate implicitly:

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

Clearly we do not need the denominator, and to minimize notation,
let us use $D = \left(x - y\right)$ and $S = \left(x + y\right)$, we need to solve :

$3 {D}^{2} S \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) - {D}^{3} \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$ for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Now divide all by ${D}^{2}$, to get:

$3 S \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) - D \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$. So, we have:

$3 S - 3 S \frac{\mathrm{dy}}{\mathrm{dx}} - D - D \frac{\mathrm{dy}}{\mathrm{dx}} = 0$. And this tells us that:

$3 S - D = \left(3 S + D\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

Return to $D = \left(x - y\right)$ and $S = \left(x + y\right)$, and simplify (divide all by 2).