# x^2y + xy = 6. Find dx/dy?

Jan 24, 2017

$\frac{\mathrm{dx}}{\mathrm{dy}} = - \frac{{x}^{2} {\left(x + 1\right)}^{2}}{6 \left(2 x + 1\right)}$.

#### Explanation:

${x}^{2} y + x y = 6$

Treating $x$ as a function of $y$, we get:

$\implies \frac{d}{\mathrm{dy}} \left({x}^{2} y\right) + \frac{d}{\mathrm{dy}} \left(x y\right) = \frac{d}{\mathrm{dy}} \left(6\right)$
$\implies \left[\frac{d \left({x}^{2}\right)}{\mathrm{dy}} \cdot y + {x}^{2} \cdot \frac{d \left(y\right)}{\mathrm{dy}}\right] + \left[\frac{d \left(x\right)}{\mathrm{dy}} \cdot y + x \cdot \frac{d \left(y\right)}{\mathrm{dy}}\right] = 0$

Let $x ' = \frac{\mathrm{dx}}{\mathrm{dy}}$.

$\implies 2 x x ' y + {x}^{2} \left(1\right) + x ' y + x \left(1\right) = 0$
$\implies 2 x x ' y + x ' y + {x}^{2} + x = 0$
$\implies x ' \left(2 x y + y\right) = \text{-} \left({x}^{2} + x\right)$
$\implies x ' = - \frac{{x}^{2} + x}{2 x y + y} = - \frac{x \left(x + 1\right)}{y \left(2 x + 1\right)}$

Also:
${x}^{2} y + x y = 6$

$\implies y \left({x}^{2} + x\right) = 6$
$\implies y = 6 {\left({x}^{2} + x\right)}^{\text{-1}} = \frac{6}{x \left(x + 1\right)}$

Substituting this into our equation for $x '$ gives us

$x ' = - \frac{x \left(x + 1\right)}{\left[\frac{6}{x \left(x + 1\right)}\right] \left(2 x + 1\right)}$

$\textcolor{w h i t e}{x '} = - \frac{{x}^{2} {\left(x + 1\right)}^{2}}{6 \left(2 x + 1\right)}$

## Another way:

We could also recognize that, since $y = \frac{6}{x \left(x + 1\right)}$, we have

dy/dx=6("-"1)(x^2+x)^"-2"(2x+1)=-(6(2x+1))/((x^2+x)^2)

And reciprocating this gives

$\frac{\mathrm{dx}}{\mathrm{dy}} = - {\left({x}^{2} + x\right)}^{2} / \left(6 \left(2 x + 1\right)\right) = - \frac{{\left[x \left(x + 1\right)\right]}^{2}}{6 \left(2 x + 1\right)} = - \frac{{x}^{2} {\left(x + 1\right)}^{2}}{6 \left(2 x + 1\right)}$

as before.

Jan 24, 2017

I got:

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

A nice trick is to recognize that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}}$. Let's try both ways (finding $\frac{\mathrm{dy}}{\mathrm{dx}}$ and taking the reciprocal, vs. finding $\frac{\mathrm{dx}}{\mathrm{dy}}$).

$f \left(x , y\right) = {x}^{2} y + x y = 6$

From the chain rule, $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$, so, using the product rule twice:

$\implies \left({x}^{2} \frac{d}{\mathrm{dy}} \left[y\right] \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] y\right) + \left(x \frac{d}{\mathrm{dy}} \left[y\right] \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \frac{d}{\mathrm{dx}} \left[x\right]\right) = 0$

$\implies \left({x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y\right) + \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right) = 0$

$\implies {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + x \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x y - y$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} \left[{x}^{2} + x\right] = - \left(2 x y + y\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x y + y}{{x}^{2} + x}$

So, $\textcolor{b l u e}{\frac{\mathrm{dx}}{\mathrm{dy}} = - \frac{{x}^{2} + x}{2 x y + y}}$.

Let's check the other way. If $x = x \left(y\right)$, then, using the product rule twice:

$\implies \left({x}^{2} \frac{d}{\mathrm{dy}} \left[y\right] + y \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] \cdot \frac{\mathrm{dx}}{\mathrm{dy}}\right) + \left(x \frac{d}{\mathrm{dy}} \left[y\right] + y \frac{d}{\mathrm{dx}} \left[x\right] \cdot \frac{\mathrm{dx}}{\mathrm{dy}}\right) = 0$

$\implies \left({x}^{2} + 2 x y \frac{\mathrm{dx}}{\mathrm{dy}}\right) + \left(x + y \frac{\mathrm{dx}}{\mathrm{dy}}\right) = 0$

$\implies 2 x y \frac{\mathrm{dx}}{\mathrm{dy}} + y \frac{\mathrm{dx}}{\mathrm{dy}} = - x - {x}^{2}$

$\implies \frac{\mathrm{dx}}{\mathrm{dy}} \left[2 x y + y\right] = - \left(x + {x}^{2}\right)$

$\implies \textcolor{b l u e}{\frac{\mathrm{dx}}{\mathrm{dy}} = - \frac{x + {x}^{2}}{2 x y + y}}$

Indeed, same result both ways.