# How do you implicitly differentiate -1=yx+(x-ye^x)/(y+x)?

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

#### Explanation:

Given equation

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

Differentiate with respect to x both sides of the equation

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

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

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

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

$0 = y + x \cdot y ' + \frac{\left(y - {y}^{2} \cdot {e}^{x} - {e}^{x} \cdot y y ' + x - x y \cdot {e}^{x} - x {e}^{x} \cdot y ' - x y ' + y {e}^{x} y ' - x + y {e}^{x}\right)}{y + x} ^ 2$

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

$0 = y {\left(y + x\right)}^{2} + x {\left(y + x\right)}^{2} y ' + y - {y}^{2} {e}^{x} - x y {e}^{x} - x {e}^{x} y ' - x y ' + y {e}^{x}$

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

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

God bless...I hope the explanation is useful.