How do you find the derivative of #f(x) = e^x + e^-x / 2 #?
1 Answer
Dec 20, 2015
Use the properties of the derivative of
#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#
