# What is the derivative of cosh(x)?

Dec 19, 2014

The definition of $\cosh \left(x\right)$ is $\frac{{e}^{x} + {e}^{-} x}{2}$, so let's take the derivative of that:

$\frac{d}{\mathrm{dx}} \left(\frac{{e}^{x} + {e}^{-} x}{2}\right)$
We can bring $\frac{1}{2}$ upfront.
$\frac{1}{2} \left(\frac{d}{\mathrm{dx}} {e}^{x} + \frac{d}{\mathrm{dx}} {e}^{-} x\right)$
For the first part, we can just use the fact that the derivative of ${e}^{x} = {e}^{x}$:
$\frac{1}{2} \left({e}^{x} + \frac{d}{\mathrm{dx}} {e}^{-} x\right)$
For the second part, we can use the same definition, but we also have to use the chain rule. For this, we need the derivative of $- x$, which is simply $- 1$:
$\frac{1}{2} \left({e}^{x} + \left(- 1\right) {e}^{-} x\right)$
$= \frac{1}{2} \left({e}^{x} - {e}^{-} x\right)$
$= \frac{{e}^{x} - {e}^{-} x}{2}$
$= \sinh \left(x\right)$ (definition of sinh).

And that's you're derivative.
Hope it helped.