# How do you find the derivative of y=(e^x+e^-x)/4?

May 24, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} - {e}^{- x}}{4} = \frac{1}{2} \sinh x$

#### Explanation:

We have:

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

Differentiating directly:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} - {e}^{- x}}{4}$

Also, If you are familiar with the hyperbolic functions then we can proceed as follows:

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

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

$\setminus \setminus = \frac{1}{2} \cosh x$

and so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \sinh x$