# What is the derivative of y=sqrt(x^2+3y^2)?

##### 1 Answer
Oct 29, 2016

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

#### Explanation:

Although you could differentiate it in it's present form using $y = {\left({x}^{2} + 3 {y}^{2}\right)}^{\frac{1}{2}}$, it is much easier to manipulate it as follows:

$y = \sqrt{{x}^{2} + 3 {y}^{2}} \implies {y}^{2} = {x}^{2} + 3 {y}^{2}$
$\therefore {x}^{2} + 2 {y}^{2} = 0$

We can now very easily differentiate this implicitly and we have the advantage of not having messy $\frac{1}{2}$ powers to deal with:

Differentiating wrt $x$ gives us:

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

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

$\therefore x + \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left({y}^{2}\right) = 0$ (this is the implicit differentiation)

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

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