# How do you use implicit differentiation to find dy/dx given x^2+3xy-y^2=0?

Oct 24, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left(3 \pm \sqrt{13}\right)$

#### Explanation:

$0 = {x}^{2} + 3 x y - {y}^{2} = \left(x + \frac{1}{2} \left(3 - \sqrt{13}\right) y\right) \left(x + \frac{1}{2} \left(3 + \sqrt{13}\right) y\right)$

so we have two solutions

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left(3 \pm \sqrt{13}\right)$

This result can be obtained the usual way

$\mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0$ so

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / \left({f}_{y}\right) = - \frac{2 x + 3 y}{3 x - 2 y}$

now substituting

$y = \frac{1}{2} \left(3 x \pm \sqrt{13} x\right)$ we obtain the same result.

This result can be also obtained by observing that this is a homogeneous equation. Making $y = \lambda x$ we obtain

${x}^{2} + 3 \lambda {x}^{2} - {\lambda}^{2} {x}^{2} = 0$ being true for all $x$ we have

$1 + 3 \lambda - {\lambda}^{2} = 0$ with solutions

$\lambda = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left(3 \pm \sqrt{13}\right)$

Oct 24, 2016

See below.

#### Explanation:

We consider $y$ an unknown function of $x$. It may help to think of it as $\text{some stuff in parentheses}$

So we have

x^2+3xunderbrace((" "))_("this stuff is called "y) - underbrace((" "))_("this is also "y) ^2 = 0

When we differentiate $3 x$ times some stuff in parentheses, we need the product rule. When we differentiate the square of some stuff in parentheses, we use the chain rule. So when we differentiate both sides of the equation (with respect to $x$), we get

${x}^{2} + 3 x y - {y}^{2} = 0$

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

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

Solve to get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x + 3 y}{2 y - 3 x}$.