# How do you implicitly differentiate 7=1-(y-x)/y^2?

Nov 12, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + 12 y}$

#### Explanation:

The equation does not express $y$ explicitly in terms of $x$ as in $y = f \left(x\right)$, instead we have $g \left(y\right) = f \left(x\right)$, so when we differentiate we apply the chain rule so that we differentiate $g \left(y\right)$ wrt $y$ rather than $x$, as in:

$g \left(y\right) = f \left(x\right)$
$\therefore \frac{d}{\mathrm{dx}} g \left(y\right) = \frac{d}{\mathrm{dx}} f \left(x\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} g \left(y\right) = f ' \left(x\right)$
$\therefore g ' \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right)$

So for $7 = 1 - \frac{y - x}{y} ^ 2$ we have:

$- 6 = \frac{y - x}{y} ^ 2$
$\therefore - 6 {y}^{2} = y - x$
$\therefore - 12 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} - 1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} + 12 y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\therefore \left(1 + 12 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + 12 y}$