# How do you differentiate 2=ye^(x-y^3)+xe^(x-y)?

$y ' = \frac{y \cdot {e}^{x - {y}^{3}} + x \cdot {e}^{x - y} + {e}^{x - y}}{3 {y}^{3} \cdot {e}^{x - {y}^{3}} - {e}^{x - {y}^{3}} + x \cdot {e}^{x - y}}$

#### Explanation:

from the given $2 = y \cdot {e}^{x - {y}^{3}} + x \cdot {e}^{x - y}$

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

$0 = y \cdot {e}^{x - {y}^{3}} \cdot \left(1 - 3 {y}^{2} \cdot y '\right) + {e}^{x - {y}^{3}} \cdot y ' + x \cdot {e}^{x - y} \cdot \left(1 - y '\right) + {e}^{x - y} \cdot 1$

Solving for $y '$ in terms of x and y then simplification

$y ' = \frac{y \cdot {e}^{x - {y}^{3}} + x \cdot {e}^{x - y} + {e}^{x - y}}{3 {y}^{3} \cdot {e}^{x - {y}^{3}} - {e}^{x - {y}^{3}} + x \cdot {e}^{x - y}}$

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