## Calculus (3rd Edition)

$1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+...$
We have the Maclaurin series for $e^x$ as follows: $$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$$ Replace $\sin x$ for $x$ in the above form. Therefore, we have: $f(x)=e^{\sin x}=1+\sin x+\dfrac{\sin^2 x}{2!}+\dfrac{\sin^3 x}{3!}+\dfrac{\sin^4 x}{4!}+...\\=1+x-\dfrac{x^3}{6}+\dfrac{1}{2}(x^2-\dfrac{x^4}{3})+\dfrac{1}{6} x^3+...\\=1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+...$