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

Dec 20, 2015

Use the properties of the derivative of ${e}^{x}$ and chain rule to find:

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

#### Explanation:

I am assuming that there's a formatting issue in the question and that you intended to write:

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

I will also assume that you know:

$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

$\frac{d}{\mathrm{dx}} \left(- x\right) = - 1$

The chain rule:

$\frac{d}{\mathrm{dx}} u \left(v \left(x\right)\right) = u ' \left(v \left(x\right)\right) \cdot v ' \left(x\right)$

Hence:

$\frac{d}{\mathrm{dx}} {e}^{- x} = {e}^{- x} \cdot \left(- 1\right) = - {e}^{- x}$

So:

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

In case you are unfamiliar with it, these are the definitions of hyperbolic cosine and hyperbolic sine:

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

$\sinh x = \frac{{e}^{x} - {e}^{- x}}{2}$

So we have shown:

$\frac{d}{\mathrm{dx}} \cosh x = \sinh x$

Similarly, we can show:

$\frac{d}{\mathrm{dx}} \sinh x = \cosh x$