# How do you implicitly differentiate x^2/16 + y^2/36 = 1?

Sep 26, 2016

Differentiate each term separately, collect the terms containing the factor $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left, move other terms to the right, and then divide both sides by the coefficient of $\frac{\mathrm{dy}}{\mathrm{dx}}$

#### Explanation:

Term 1:

$\frac{d \left({x}^{2} / 16\right)}{\mathrm{dx}} = 2 \frac{x}{16} = \frac{x}{8}$

Term 2:

$\frac{d \left({y}^{2} / 36\right)}{\mathrm{dx}} = \left(2 \frac{y}{36}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{y}{18} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Term 3;

$\frac{d \left(1\right)}{\mathrm{dx}} = 0$

Returning the 3 terms to their place in the equation:

$\frac{x}{8} + \frac{y}{18} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

Collect terms that contain $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left and move any other terms to the right:

$\frac{y}{18} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - \frac{x}{8}$

Divide by the coefficient of $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- \frac{x}{8}\right) \left(\frac{18}{y}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 9 x}{4 y}$