# How do you differentiate 1=e^(xy)/(e^x+xy)?

##### 1 Answer

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

#### Explanation:

From the given

$1 = {e}^{x y} / \left({e}^{x} + x y\right)$

${e}^{x} + x y = {e}^{x y}$

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

${e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(x\right) + x \cdot \frac{d}{\mathrm{dx}} \left(y\right) + y \cdot \frac{d}{\mathrm{dx}} \left(x\right) = {e}^{x y} \cdot \left[x \cdot \frac{d}{\mathrm{dx}} \left(y\right) + y \cdot \frac{d}{\mathrm{dx}} \left(x\right)\right]$

${e}^{x} + x y ' + y \cdot 1 = {e}^{x y} \cdot \left(x y ' + y \cdot 1\right)$

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

${e}^{x} + y - y \cdot {e}^{x y} = x y ' \cdot {e}^{x y} - x y '$

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

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

But, ${e}^{x y} = {e}^{x} + x y$

therefore

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

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

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