Use multiplication of Taylor series to find the quartic Taylor polynomial about 0 for the function:

\(\displaystyle f(x)=\frac{sinx}{\sqrt{1+x}}\) evaluating the coefficients.

For sin x the standard Taylor series about 0 is:

\(\displaystyle x-\frac{1}{3!}x^3+...\)

For \(\displaystyle \sqrt{1+x}\) this can be rearranged to fit the standard Talor series

\(\displaystyle (1+x)^\alpha\), which is

\(\displaystyle 1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3\) ...

Multiplying these two together the answer I get is

\(\displaystyle \frac{1}{2}x+\frac{3}{8}x^2+\frac{5}{96}x^3+\frac{35}{128}x^4\)

Does this look along the right lines? (Worried)