# How do you implicitly differentiate  2x+2y = sqrt(x^2+y^2)?

##### 1 Answer
Jun 15, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left\{- \frac{1}{3} \left(\sqrt{7} + 4\right) , \frac{1}{3} \left(\sqrt{7} - 4\right)\right\}$

depending on de sign of $x$

#### Explanation:

$f \left(x , y\right) = 2 x + 2 y - \sqrt{{x}^{2} + {y}^{2}} = 0$ is an homogeneus function.

Making the substitution

$y = \lambda x$ we get at

$f \left(x , \lambda x\right) = 2 x \left(1 + \lambda\right) - \left\mid x \right\mid \sqrt{1 + {\lambda}^{2}} = 0$ or

$\frac{x}{\left\mid x \right\mid} = \frac{1}{2} \frac{\sqrt{1 + {\lambda}^{2}}}{1 + \lambda} = \pm 1$

Solving for $\lambda$ we have

$\frac{x}{\left\mid x \right\mid} = 1 \to \lambda = \frac{1}{3} \left(\sqrt{7} - 4\right)$
$\frac{x}{\left\mid x \right\mid} = - 1 \to \lambda = - \frac{1}{3} \left(\sqrt{7} + 4\right)$

so $\frac{\mathrm{dy}}{\mathrm{dx}} = \left\{- \frac{1}{3} \left(\sqrt{7} + 4\right) , \frac{1}{3} \left(\sqrt{7} - 4\right)\right\}$

depending on de sign of $x$