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

Jun 17, 2017

Given: ${x}^{2} + 3 x y + {y}^{2} = 0$

Differentiate each term with respect to x:

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

Use the power rule, $\frac{\mathrm{dy}}{\mathrm{dx}} = n {x}^{x - 1}$, on the first term:

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

Use the product rule, $\frac{d \left(x y\right)}{\mathrm{dx}} = \frac{\mathrm{dx}}{\mathrm{dx}} y + x \frac{\mathrm{dy}}{\mathrm{dx}} = y + x \frac{\mathrm{dy}}{\mathrm{dx}}$ on the second term:

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

Use the chain rule, $\frac{d \left({y}^{2}\right)}{\mathrm{dx}} = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$, on the third term:

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

The derivative of a constant is 0:

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

Distribute the 3:

$2 x + 3 y + 3 x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Move all of the terms that do not contain $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the right:

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

Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

Divide by $3 x + 2 y$:

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

Jun 17, 2017

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

#### Explanation:

$\text{differentiate "color(blue)"implicitly with respect to x}$

$\text{the term " 3xy" is differentiated using the "color(blue)"product rule}$

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

$\Rightarrow 2 x + 3 x \frac{\mathrm{dy}}{\mathrm{dx}} + 3 y + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

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

Jun 17, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \left(\sqrt{5} \pm 3\right)$

#### Explanation:

From ${x}^{2} + 3 x y + {y}^{2} = 0 \to y = - \frac{1}{2} \left(\sqrt{5} \pm 3\right) x$

then

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \left(\sqrt{5} \pm 3\right)$