# How do you differentiate (xy)/(x+y)=1?

Jan 20, 2017

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

#### Explanation:

Using the quotient rule

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v u ' - u v '}{v} ^ 2$

$\frac{d}{\mathrm{dx}} \left(\frac{x y}{x + y} = 1\right)$

$\frac{\left(x + y\right) \left(y + x y '\right) - \left(x y \left(1 + y '\right)\right)}{x + y} ^ 2 = 0$

ie.$\left(x + y\right) \left(y + x y '\right) - x y \left(1 + y '\right) = 0$

$\cancel{x y} + {x}^{2} y ' + {y}^{2} + \cancel{x y y '} \cancel{- x y} - \cancel{x y y '} = 0$

${x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2} = 0$

giving

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

Jan 22, 2017

We can also rewrite the function from the outset to avoid fractions:

$x y = x + y$

Then we see that the right-hand side will use the quotient rule: $\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$. Recall that differentiating anything with $y$ will cause $\frac{\mathrm{dy}}{\mathrm{dx}}$ to spit out thanks to the chain rule.

Differentiating gives:

$\left[\frac{d}{\mathrm{dx}} x\right] y + x \left[\frac{d}{\mathrm{dx}} y\right] = \left[\frac{d}{\mathrm{dx}} x\right] + \left[\frac{d}{\mathrm{dx}} y\right]$

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

Grouping the $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms:

$x \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - y$

Factoring:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(x - 1\right) = 1 - y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - y}{x - 1}$

Jan 22, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x - 1} ^ 2$.

#### Explanation:

We have, $\frac{x y}{x + y} = 1 \Rightarrow x y = x + y \Rightarrow x y - y = x$

$\therefore y \left(x - 1\right) = x \Rightarrow y = \frac{x}{x - 1} , x \ne 1$

:. y'={(x-1)(x)'-(x)(x-1)'}/(x-1)^2............[because, "the Quotient Rule]"

$= \frac{\left(x - 1\right) \left(1\right) - \left(x\right) \left(1 - 0\right)}{x - 1} ^ 2 = - \frac{1}{x - 1} ^ 2$

$\text{Therefore, } \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x - 1} ^ 2$.