Question

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

Answer 1

Use the properties of the derivative of `e^x` and chain rule to find:

`d/(dx) ((e^x+e^(-x))/2) = (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(x) = (e^x+e^(-x))/2`

I will also assume that you know:

`d/(dx) e^x = e^x`

`d/(dx) (-x) = -1`

The chain rule:

`d/(dx) u(v(x)) = u'(v(x))*v'(x)`

Hence:

`d/(dx) e^(-x) = e^(-x)*(-1) = -e^(-x)`

So:

`d/(dx) ((e^x+e^(-x))/2) = (e^x-e^(-x))/2`

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

`cosh x = (e^x+e^(-x))/2`

`sinh x = (e^x-e^(-x))/2`

So we have shown:

`d/(dx) cosh x = sinh x`

Similarly, we can show:

`d/(dx) sinh x = cosh x`