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

1 Answer
Dec 19, 2014

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.