Question

What is the derivative of `cosh(x)`?

Answer 1

The definition of `cosh(x)` is `(e^x + e^-x)/2`, so let's take the derivative of that:

`d/dx ((e^x + e^-x)/2)`
We can bring `1/2` upfront.
`1/2(d/dxe^x + d/dxe^-x)`
For the first part, we can just use the fact that the derivative of `e^x = e^x`:
`1/2(e^x + d/dxe^-x)`
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`:
`1/2(e^x + (-1)e^-x)`
`=1/2(e^x - e^-x)`
`=(e^x-e^-x)/2`
`=sinh(x)` (definition of sinh).

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