# Question 12cad

Oct 9, 2016

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

#### Explanation:

Term 1:

Use the chain rule:

(d(y²))/dx = 2ydy/dx#

Term 2:

Use the chain rule:

$\frac{d \left\{f \left(u \left(x\right)\right)\right\}}{\mathrm{dx}} = \left(\frac{\mathrm{df} \left(u\right)}{\mathrm{du}}\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

let $u \left(x\right) = - x y$, then $f \left(u\right) = 2 {e}^{u}$

$\frac{\mathrm{df} \left(u\right)}{\mathrm{du}} = \frac{d \left(2 {e}^{u}\right)}{\mathrm{du}} = 2 {e}^{u}$

Use the product rule on u to find $\frac{\mathrm{du}}{\mathrm{dx}}$:

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

Substituting into the chain rule:

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

Reverse the substitution for u:

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

Term 3:

$\frac{d 6}{\mathrm{dx}} = 0$

Return the terms back into the equation:

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

Move the terms not containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the right side:

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

Factor out 2 and $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left:

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

Divide by the coefficient of $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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