# How do you differentiate f(x)= 2 / (e^x + e^-x)^3?

Oct 4, 2016

$= - 6 \frac{{e}^{x} - {e}^{- x}}{{\left({e}^{x} + {e}^{-} x\right)}^{4}}$

#### Explanation:

$f \left(x\right) = \frac{2}{{e}^{x} + {e}^{-} x} ^ 3$

$= \frac{2}{2 \left(\cosh x\right)} ^ 3 = \frac{1}{4} {\sech}^{3} x$

$f ' \left(x\right) = 3 \left(\frac{1}{4}\right) {\sech}^{2} x \left(- \sech x\right) \tanh x$

$= - \frac{3}{4} {\sech}^{3} x \tanh x$

$= - \frac{3}{4} \frac{1}{\frac{{e}^{x} + {e}^{-} x}{2}} ^ 3 \cdot \frac{{e}^{x} - {e}^{- x}}{{e}^{x} + {e}^{- x}}$

$= - 6 \frac{{e}^{x} - {e}^{- x}}{{\left({e}^{x} + {e}^{-} x\right)}^{4}}$