How do I prove that the differential of coshx is equal to sinhx?

Before this proof, it is important to appreciate that both of these hyperbolic  functions can be written in terms of e^x. Therefore, before you begin to differentiate, you must represent coshx as (e^x + e^-x)/2. Then, this can easily be differentiated to give you the answer (e^x-e^-x)/2, which is the equivalent of sinhx.