# How do you differentiate y = 2/(e^(x) + e^(-x))?

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