# Question #0e708

Nov 26, 2016

$\frac{d}{\mathrm{dx}} \frac{{e}^{x} - {e}^{-} x}{{e}^{x} + {e}^{-} x} = \frac{4}{{e}^{x} + {e}^{-} x} ^ 2$

#### Explanation:

Using the quotient rule and the sum rule along with the following:

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

we have

$\frac{d}{\mathrm{dx}} \frac{{e}^{x} - {e}^{-} x}{{e}^{x} + {e}^{-} x}$

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

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

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

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

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

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