# What is the implicit derivative of 5=y/(x-y)?

##### 1 Answer
Nov 20, 2015

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

#### Explanation:

Quotient Rule for Derivatives:
$d \frac{\frac{u}{v}}{\mathrm{dx}} = \frac{v \cdot \frac{\mathrm{du}}{\mathrm{dx}} - u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$

With $u = y$ and $v = \left(x - y\right)$
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$
and
$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{d \left(x - y\right)}{\mathrm{dx}} = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$

Therefore
$\textcolor{w h i t e}{\text{XXX}} \frac{d \left(\frac{y}{x - y}\right)}{\mathrm{dx}} = \frac{\left(x - y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - y \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{\left(x - y\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXXXXXXX}} = \frac{x \frac{\mathrm{dy}}{\mathrm{dx}} - y}{{\left(x - y\right)}^{2}}$

$\frac{d \left(5\right)}{\mathrm{dx}} = 0$

So, since $5 = \frac{y}{x - y}$

$\Rightarrow \textcolor{w h i t e}{\text{XXX}} \frac{d \left(5\right)}{\mathrm{dx}} = \frac{d \left(\frac{y}{x - y}\right)}{\mathrm{dx}}$

$\Rightarrow \textcolor{w h i t e}{\text{XXX}} 0 = \frac{x \frac{\mathrm{dy}}{\mathrm{dx}} - y}{{\left(x - y\right)}^{2}}$

$\Rightarrow \textcolor{w h i t e}{\text{XXX}} x \frac{\mathrm{dy}}{\mathrm{dx}} - y = 0$

$\Rightarrow \textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x}$