Question

y=e^x+e^-x

Answer 1

And the question is...?

Answer 2

Use the series expansions of both terms and add:
\[e ^{x}=1+x+\frac{x ^{2}}{2!}+\frac{x ^{3}}{3!}+\frac{x ^{4}}{4!}\ .............\]
\[e ^{-x}=1-x+\frac{x ^{2}}{2!}-\frac{x ^{3}}{3!}+\frac{x ^{4}}{4!}\ .............\]