First $3$ non-zero terms of the Maclaurin Series $\frac{1}{\sqrt{4+x^3}}$ Since each derivative will be multiplied by $3x^2$, are all the terms of this Maclaurin series $0$?
 A: To find the Maclaurin series, first find the series for
$$f(x) = {1\over \sqrt{4 + x}}$$
then do the substitution $x \leftarrow x^3$.
A: You can use the generalized binomial theorem

$$ (4+x^3)^{-1/2}= \frac{1}{2}(1+x^3/4)^{-1/2} = \frac{1}{2}\sum_{k=0}^{\infty} {-1/2 \choose k } \frac{x^{3k}}{4^k}, $$

where 

$$ {-1/2 \choose k } =\frac{(-1/2)!}{k!(-1/2-k)!}. $$

A: Your answer lies within the question, it states to find the first three non-zero terms. It is a trick question as the first two derivatives are indeed zero at $f'(0)$ and $f''(0)$ but keep differentiating and don't forget to use the product rule and you shall be rewarded with an answer for your efforts.
Oh I am still new here and unfortunately don't have enough rep to comment yet, so I apologise as this next statement should be a comment:
@Mhenni do you think $(-1/2)!$ should be written as $\Gamma\left(\frac12\right)=\sqrt\pi$ to avoid the confusion of negative factorials maybe with a link to the Gamma Function that generalises the factorial notation?
