# How do you find the derivative of exponential function f(x)= 2 / (e^x + e^-x)^3?

Apr 10, 2018

$\frac{d}{\mathrm{dx}} \left(\frac{2}{{e}^{x} + {e}^{-} x} ^ 3\right) = - \frac{6 \left({e}^{x} - {e}^{-} x\right)}{{e}^{x} + {e}^{-} x} ^ 4 = - \frac{3 \sinh x}{4 {\cosh}^{4} x}$

#### Explanation:

Using the chain rule:

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

$\frac{d}{\mathrm{dx}} \left(\frac{2}{{e}^{x} + {e}^{-} x} ^ 3\right) = 2 \left(- 3\right) {\left({e}^{x} + {e}^{-} x\right)}^{-} 4 \frac{d}{\mathrm{dx}} \left({e}^{x} + {e}^{-} x\right)$

$\frac{d}{\mathrm{dx}} \left(\frac{2}{{e}^{x} + {e}^{-} x} ^ 3\right) = - 6 {\left({e}^{x} + {e}^{-} x\right)}^{-} 4 \left({e}^{x} - {e}^{-} x\right)$

$\frac{d}{\mathrm{dx}} \left(\frac{2}{{e}^{x} + {e}^{-} x} ^ 3\right) = - \frac{6 \left({e}^{x} - {e}^{-} x\right)}{{e}^{x} + {e}^{-} x} ^ 4$

Note also that:

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

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

$\frac{d}{\mathrm{dx}} \left(\frac{2}{{e}^{x} + {e}^{-} x} ^ 3\right) = - \frac{3 \sinh x}{4 {\cosh}^{4} x}$