# How do you implicitly differentiate  (x^2)/9+(x-y^2)/6=1 ?

Mar 11, 2016

I found: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{y} \left(\frac{4 x + 3}{6}\right)$

#### Explanation:

In this implicit expression we do not isolate $y$ but we leave it as a function of $x$; when we differentiate we also need to differentiate $y$ as well with respect to $x$. So, for example, if you have ${y}^{2}$ differentiating you get: $2 y \frac{\mathrm{dy}}{\mathrm{dx}}$.
In our case we get:
$\frac{1}{9} \cdot 2 x + \frac{1}{6} \left(1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$\frac{2}{9} x + \frac{1}{6} - \frac{1}{3} y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
collect $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{y} \left(\frac{2}{9} x + \frac{1}{6}\right)$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{y} \left(\frac{2}{3} x + \frac{1}{2}\right) = \frac{1}{y} \left(\frac{4 x + 3}{6}\right)$.