Maclaurin Series with $f^{(n)}(0)=0$

Question

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 $(4+x^2)^{\frac{-1}{2}}$, proper form of binomial function is $\frac{1}{2}(1+\frac{x^2}{4})^{\frac{-1}{2}}$,

the first derivative becomes $\frac{-1}{2}(1+\frac{x^2}{4})^{\frac{-3}{2}}*\frac{x}{2}$ and every following derivatives have $\frac{x}{2}$ 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 $\frac{x}{2}$

Answer 1

No

$f'(x)=-\dfrac{x}{{{\left( {{x}^{2}}+4\right) }^{3/2}}}$ and so indeed $f'(0)=0$

But $f''(x)=\dfrac{3 {{x}^{2}}}{{{\left( {{x}^{2}}+4\right) }^{5/2}}}-\dfrac{1}{{{\left( {{x}^{2}}+4\right) }^{3/2}}}$ so $f''(0)=0-\dfrac18$ and something similar happens with all the even derivatives

In fact $${{{\left( {{x}^{2}}+4\right) }^{-1/2}}}= \frac{1}{2}-\frac{1}{16} {{x}^{2}}+\frac{3}{256} {{x}^{4}}-\frac{5}{2048} {{x}^{6}}+\frac{35}{65536} {{x}^{8}}-\cdots$$

Answer 2

No, not every derivative has an $x/2$. In your example, $$f^{(2)}(x) = {\frac {3\,{x}^{2}}{32\, \left( 1+{x}^{2}/4 \right) ^{5/2}}}-1/8\, \left( 1+{x}^{2}/4 \right) ^{-3/2} $$ so $f^{(2)}(0) = -1/8$.