# How do you implicitly differentiate ln(x-y)=x/y ?

##### 2 Answers
Jan 17, 2018

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

#### Explanation:

$L n \left(x - y\right) = \frac{x}{y}$

$\frac{1 - y '}{x - y} = \frac{1 \cdot y - x y '}{x} ^ 2$

$\frac{1 - y '}{x - y} = \frac{y - x y '}{x} ^ 2$

#(1-y')x^2=(x-y)(y-xy')

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

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

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

Jan 17, 2018

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

#### Explanation:

$\text{differentiate "ln(x-y)" using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\ln \left(x - y\right)\right)$

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

$\text{differentiate "x/y" using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\frac{x}{y}\right) = \frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2$

$\text{equating the derivates on both sides}$

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{2} - x y - {y}^{2}\right) = x y - 2 {y}^{2}$

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