# What is the implicit derivative of 1= x/y-e^(xy) ?

Apr 1, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - {e}^{x y} {y}^{3}}{x - x {e}^{x y} {y}^{2}}$

#### Explanation:

$1 = \frac{x}{y} - {e}^{x y}$

First we have to know that we can differentiate each part separately

Take $y = 2 x + 3$ we can differentiate $2 x$ and $3$ seperately

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} 2 x + \frac{\mathrm{dy}}{\mathrm{dx}} 3 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 2 + 0$

So similarly we can differentiate $1$, $\frac{x}{y}$ and ${e}^{x y}$ separately

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

Rule 1: $\frac{\mathrm{dy}}{\mathrm{dx}} C \Rightarrow 0$ derivative of a constant is 0

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \frac{x}{y}$ we have to differentiate this using the quotient rule

Rule 2: $\frac{\mathrm{dy}}{\mathrm{dx}} \frac{u}{v} \Rightarrow \frac{\frac{\mathrm{du}}{\mathrm{dx}} v - \frac{\mathrm{dv}}{\mathrm{dx}} u}{v} ^ 2$ or $\frac{v u ' - u v '}{v} ^ 2$

$u = x \Rightarrow u ' = 1$

Rule 2: ${y}^{n} \Rightarrow \left(n {y}^{n - 1} \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$v = y \Rightarrow v ' = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{v u ' + u v '}{v} ^ 2 = \frac{1 y - \frac{\mathrm{dy}}{\mathrm{dx}} x}{y} ^ 2$

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

Lastly we have to differentiate ${e}^{x y}$ using a mixture of the chain and the product rule

Rule 3: ${e}^{u} \Rightarrow u ' {e}^{u}$

So in this case $u = x y$ which is a product

Rule 4: $\frac{\mathrm{dy}}{\mathrm{dx}} x y = y ' x + x ' y$

$x \Rightarrow 1$

$y \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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

Expand out

$0 = \frac{1 y - \frac{\mathrm{dy}}{\mathrm{dx}} x}{y} ^ 2 - \frac{\mathrm{dy}}{\mathrm{dx}} x {e}^{x y} + y {e}^{x y}$

Times both sides by ${y}^{2}$

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

$0 = y - \frac{\mathrm{dy}}{\mathrm{dx}} x - \frac{\mathrm{dy}}{\mathrm{dx}} x {e}^{x y} {y}^{2} + {e}^{x y} {y}^{3}$

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

$y - {e}^{x y} {y}^{3} = \frac{\mathrm{dy}}{\mathrm{dx}} x - \frac{\mathrm{dy}}{\mathrm{dx}} x {e}^{x y} {y}^{2}$

Factorize out $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the RHS (right hand side)

$- y - {e}^{x y} {y}^{3} = \frac{\mathrm{dy}}{\mathrm{dx}} \left(x - x {e}^{x y} {y}^{2}\right)$

$\frac{- \left(y + {e}^{x y} {y}^{3}\right)}{x - x {e}^{x y} {y}^{2}} = \frac{\mathrm{dy}}{\mathrm{dx}}$