# How do you implicitly differentiate 4=y-(x-e^y)/(y-x)?

Mar 31, 2018

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

#### Explanation:

First we have to familrise ourselves with some calculs rules

$f \left(x\right) = 2 x + 4$ we can differentiate $2 x$ and $4$ separately

$f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} 2 x + \frac{\mathrm{dy}}{\mathrm{dx}} 4 = 2 + 0 = 2$

Similarly we can differentiate the $4$, $y$ and $- \frac{x - {e}^{y}}{y - x}$ separately

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

We know that differentiating constants $\frac{\mathrm{dy}}{\mathrm{dx}} 4 = 0$

$0 = \frac{\mathrm{dy}}{\mathrm{dx}} y - \frac{\mathrm{dy}}{\mathrm{dx}} \frac{x - {e}^{y}}{y - x}$

Likewise the rule for differentiating y is $\frac{\mathrm{dy}}{\mathrm{dx}} y = \frac{\mathrm{dy}}{\mathrm{dx}}$

$0 = \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} \frac{x - {e}^{y}}{y - x}$

Lastly to differentiate $\frac{x - {e}^{y}}{y - x}$ we have to use the quotient rule

Let $x - {e}^{y} = u$
and
Let $y - x = v$

The quotient rule is $\frac{v u ' - u v '}{v} ^ 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} x - \frac{\mathrm{du}}{\mathrm{dx}} {e}^{y}$

When deriving e we use the chain rule such that ${e}^{y} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} {e}^{y}$

so $u ' = 1 - \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y}$

$y - x = v$

so

$v ' = \frac{\mathrm{dv}}{\mathrm{dx}} y - \frac{\mathrm{dv}}{\mathrm{dx}} x$

Using the same rules from above it becomes

$v ' = \frac{\mathrm{dy}}{\mathrm{dx}} - 1$

Now we have to do the quotient rule

$\frac{v u ' - u v '}{v} ^ 2 = \frac{\left(y - x\right) \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y}\right) - \left(x - {e}^{y}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)}{y - x} ^ 2$

$0 = \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\left(y - x\right) \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y}\right) - \left(x - {e}^{y}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)}{y - x} ^ 2$

Expand out

$0 = \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\left(y - y \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y} - x + x \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y}\right) - \left(x \frac{\mathrm{dy}}{\mathrm{dx}} - x - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y}\right)}{y - x} ^ 2$

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

Multiply both sides by (y-x)^2

$0 = \frac{\mathrm{dy}}{\mathrm{dx}} {\left(y - x\right)}^{2} - \left(y - y \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y} + x \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y} - x \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y}\right)$

$0 = \frac{\mathrm{dy}}{\mathrm{dx}} {\left(y - x\right)}^{2} - y + y \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y} - x \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y} + x \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y}$

Place all the $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms on one side

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

Factories dy/dx out of every term

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

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

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