# How do you find the first and second derivative of y=e^x+e^-x?

Mar 2, 2017

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

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

#### Explanation:

Using the standard result:

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

Then if :

$y = {e}^{x} + {e}^{- x}$

We have:

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

And then:

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