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

Proof: It is helpful to note that #sinh(x):=(e^x-e^-x)/2# and #cosh(x):=(e^x+e^-x)/2#. We can differentiate from here using either the quotient rule or the sum rule. I'll use the sum rule first:

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

#= (e^x)/2-(e^-x)/2#

#=>(d(sinh(x)))/dx = d/dx((e^x)/2-(e^-x)/2)#

#=d/dx(e^x/2)-d/dx(e^-x/2)# by the sum rule

#=e^x/2-(-e^x/2)# by basic differentiation of exponential functions

#=(e^x+e^-x)/2 = cosh(x).#

The quotient rule is just as easy:

Let #u=e^x-e^-x# and #v=2#

Hence #(du)/dx=e^x+e^-x# (by the sum rule and basic differentiation of exponential functions)

and #(dv)/dx=0#

Recalling that #(d(u/v))/dx = (v((du)/dx)-u((dv)/dx))/v^2#

so #(d(sinh(x)))/dx=(2*(e^x+e^-x)-0)/2^2#

#=(2*(e^x+e^-x))/4#

#=(e^x+e^-x)/2=cosh(x).#