# How do you find (dy)/(dx) given x^2+y^2=1?

##### 1 Answer
Oct 11, 2016

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

#### Explanation:

${x}^{2} + {y}^{2} = 1$

Differentiate wrt $x$:
$\frac{d}{\mathrm{dx}} {x}^{2} + \frac{d}{\mathrm{dx}} {y}^{2} = \frac{d}{\mathrm{dx}} 1$

We already know how to deal with the first and third terms, so lets get them out the way:
$\frac{d}{\mathrm{dx}} {x}^{2} + \frac{d}{\mathrm{dx}} {y}^{2} = \frac{d}{\mathrm{dx}} 1$
$\therefore 2 x + \frac{d}{\mathrm{dx}} {y}^{2} = 0$

For the remaining term we use the chain rule, we don't know how to differentiate ${y}^{2}$ wrt $x$ but we do know how to differentiate ${y}^{2}$ wrt $y$ (it the same as differentiating ${x}^{2}$ wrt $x$!). The chain rule tells us that:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$, so we can rewrite:

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

We can know perform that final differentiation, as we are now differentiating a function of $y$ wrt $y$ so
$2 x + \frac{d}{\mathrm{dy}} \left({y}^{2}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 2 x + 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
We can then rearrange to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ as follows:
$2 x + 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x$

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

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