# What is the derivative of (e^x-e^-x)/(e^x+e^-x)?

Aug 6, 2015

${y}^{'} = \frac{4 {e}^{2 x}}{{e}^{2 x} + 1} ^ 2$

#### Explanation:

You can differentiate this function by using the quotient rule and the chain rule.

Before using these rules to differentiate the function, try to simplify it a little - this will save you a lot of work on the derivative. For example, you could write

$y = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{x}}}} \cdot \left(1 - {e}^{- 2 x}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{x}}}} \cdot \left(1 + {e}^{- 2 x}\right)} = \left(1 - \frac{1}{e} ^ \left(2 x\right)\right) \cdot \left(\frac{1}{1 + \frac{1}{e} ^ \left(2 x\right)}\right)$

$y = \frac{{e}^{2 x} - 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2} x}}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{2} x}}}}{{e}^{2 x} + 1} = \frac{{e}^{2 x} - 1}{{e}^{2 x} + 1}$

So, you know that you can differentiate a function that takes the form

$y = f \frac{x}{g} \left(x\right)$, with $g \left(x\right) \ne 0$

by using the formula

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\left[\frac{d}{\mathrm{dx}} f \left(x\right)\right] \cdot g \left(x\right) - f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}{g \left(x\right)} ^ 2}$

In your case, you have $f \left(x\right) = {e}^{2 x} - 1$ and $g \left(x\right) = {e}^{2 x} + 1$, so you can write

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left({e}^{2 x} - 1\right)\right] \cdot \left({e}^{2 x} + 1\right) - \left({e}^{2 x} - 1\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{2 x} + 1\right)}{{e}^{2 x} + 1} ^ 2$

y^' = (e^(2x) * 2 * (e^(2x) + 1) - (e^(2x)-1) * e^(2x) * 2)/((e^(2x) + 1)^2

${y}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {e}^{4} x}}} + 2 \cdot {e}^{2 x} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{e}^{4} x}}} + 2 \cdot {e}^{2 x}}{{e}^{2 x} + 1} ^ 2$

${y}^{'} = \textcolor{g r e e n}{\frac{4 {e}^{2 x}}{{e}^{2 x} + 1} ^ 2}$

Aug 27, 2015

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

#### Explanation:

Hyperbolic function:
$\sinh z = \frac{1}{2} \left({e}^{z} - {e}^{- z}\right)$
$\cosh z = \frac{1}{2} \left({e}^{z} + {e}^{- z}\right)$
$\tanh z = \sinh \frac{z}{\cosh} z$
$\sech z = \frac{1}{\cosh} z$

$\frac{d}{\mathrm{dx}} \left(\frac{{e}^{x} - {e}^{- x}}{{e}^{x} + {e}^{- x}}\right) = \frac{d}{\mathrm{dx}} \left(\tanh x\right) = {\sech}^{2} x = {\left(\frac{2}{{e}^{x} + {e}^{- x}}\right)}^{2} = \frac{4 {e}^{2 x}}{{e}^{2 x} + 1} ^ 2$