# How do you implicitly differentiate -y=xy-xe^-y ?

Feb 2, 2017

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

#### Explanation:

Since, the Problem is to be solved using Implicit Diffn., we will

first solve it accordingly.

$- y = x y - x {e}^{-} y \Rightarrow x {e}^{-} y = x y + y$

$\therefore \frac{d}{\mathrm{dx}} \left(x {e}^{-} y\right) = \frac{d}{\mathrm{dx}} \left(x y + y\right) = \frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left(y\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \text{[Addition Rule]}$

By the Product Rule , then,

$x \frac{d}{\mathrm{dx}} {e}^{-} y + {e}^{-} y \frac{d}{\mathrm{dx}} \left(x\right) = \left\{x \frac{d}{\mathrm{dx}} \left(y\right) + y \frac{d}{\mathrm{dx}} \left(x\right)\right\} + \frac{\mathrm{dy}}{\mathrm{dx}} ,$ i.e.,

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

To find $\frac{d}{\mathrm{dx}} \left({e}^{-} y\right) ,$ we use the Chain Rule to get,

$\frac{d}{\mathrm{dx}} \left({e}^{-} y\right) = \frac{d}{\mathrm{dy}} \left({e}^{-} y\right) \frac{d}{\mathrm{dx}} \left(- y\right) = \left({e}^{-} y\right) \left(- \frac{\mathrm{dy}}{\mathrm{dx}}\right) .$

Hence, continuing from $\left(\star\right)$,

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{-} y - y}{x {e}^{-} y + x + 1} , \text{ &, as } {e}^{-} y = \frac{x y + y}{x} ,$ we have,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{x y + y}{x} - y}{x y + y + x + 1} = \frac{y}{x \left\{y \left(x + 1\right) + 1 \left(x + 1\right)\right\}} , i . e . ,$

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

Enjoy Maths.!

Feb 2, 2017

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

#### Explanation:

As a Second Method , we can solve the Problem without the

use of Implicit Diffn.

Let us rewrite the given eqn. as, $y = x {e}^{-} y - x y = x \left({e}^{-} y - y\right) ,$

giving, $x = \frac{y}{{e}^{-} y - y} .$

Diff.ing w.r.t. $y$ using the Quotient Rule and the Chain Rule,

$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{\left({e}^{-} y - y\right) \frac{d}{\mathrm{dy}} \left(y\right) - y \frac{d}{\mathrm{dy}} \left({e}^{-} y - y\right)}{{e}^{-} y - y} ^ 2 ,$ i.e.,

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

$= \frac{{e}^{-} y - y + y {e}^{-} y + y}{{e}^{-} y - y} ^ 2$

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

$\text{Therefore, } \frac{\mathrm{dy}}{\mathrm{dx}} = {\left({e}^{-} y - y\right)}^{2} / \left(\left(y + 1\right) {e}^{-} y\right) .$

I leave it, as an exercise, to show that our both Answers match!