Why do we need to reindex Taylor Series when differentiating cosh

Question

I'm doing a calculus course right now and I do not quite understand why the Taylor series needs to be reindexed when we take the derivative of cosh (second row in the picture). Can someone please explain?

The picture with sinh and cosh derivatives in Taylor series form

Answer 1

You could just as well not reindex it, but the $k=0$ term is $$(2k)\frac{x^{2k-1}}{(2k)!}=0$$ You can see directly that the derivative of that term is $0$ because it is the constant $1$. Therefore reindexing it is cleaner as it leaves out unnecessary terms.