Vtyjkyuk

I am learning Maclaurin Series for the first time and having trouble understanding it.

The thing is, Maclaurin Series has the basic thinking that infinite number of derivatives have coefficients of $f^(n)(0)$ that equals $n!C_n$. I get that.

But whenever there is a chain rule, does every $f^(n)(0)$ not become 0?

For example, for function $(4+x^2)^frac-12$, proper form of binomial function is
$frac12(1+fracx^24)^frac-12$,

the first derivative becomes $frac-12(1+fracx^24)^frac-32*fracx2$
and every following derivatives have $fracx2$ at the end, which makes $f^(n)(0)=0$ because x being zero makes everything zero.

But the solutions don't seem to mind it and solve problems as if there was not $fracx2$

No

$f'(x)=-dfracxleft( x^2+4right) ^3/2$ and so indeed $f'(0)=0$

But $f''(x)=dfrac3 x^2left( x^2+4right) ^5/2-dfrac1left( x^2+4right) ^3/2$ so $f''(0)=0-dfrac18$ and something similar happens with all the even derivatives

In fact $$left( x^2+4right) ^-1/2= frac12-frac116 x^2+frac3256 x^4-frac52048 x^6+frac3565536 x^8-cdots$$

No, not every derivative has an $x/2$. In your example,
$$f^(2)(x) = frac 3,x^232, left( 1+x^2/4 right) ^5/2-1/8,
left( 1+x^2/4 right) ^-3/2
$$ so $f^(2)(0) = -1/8$.