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

1 Answer
Dec 20, 2015

Answer:

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#