Maclaurin's series for $\cos{\sqrt{x}}$ Why does Maclaurin's series for this function exist:
$$y=\cos{\sqrt{x}}$$
even though the second derivative of the function at $x=0$ is undefined?
I know that you can use the standard series of $\cos{x}$ and replacing $x$ with $\sqrt{x}$ to find its Maclaurin's expansion, but why can't I use the standard Maclaurin's theorem to expand the function as a power series?

You seem to misinterpret the nature of the chain rule $$(f\circ g)'(x) = f'(g(x))g'(x).$$ What it means is that if $g$ is differentiable at $x_0$ and $f$ is differentiable at $g(x_0)$, then $f\circ g$ is differentiable at $x_0$ and the derivative is given by the above formula.
Basically, a crude way to say it is "conditions xyz are true $\implies$ formula works" and its contrapositive is "formula doesn't work $\implies$ conditions xyz are not true". The contrapositive of the chain rule doesn't say that composition isn't differentiable if the formula doesn't work.
For example, $(\cos\sqrt x)' = -\frac{\sin\sqrt x}{2\sqrt x}$ for $x\neq 0$ by the chain rule. As you say, it is not defined at $0$, so the chain rule doesn't work at $0$ and from that we can conclude that either $\cos$ or $\sqrt{\,\cdot\,}$ is not differentiable at $0$. And take a look at that, indeed we know that $\sqrt{\,\cdot\,}$ is not differentiable at $0$.
However, that still doesn't mean that $\cos\sqrt x$ isn't differentiable at $0$. We can check the definition of the derivative, in this case we must verify that the limit $$\lim_{x\to 0}\frac{\cos\sqrt x - 1}{x}$$ exists. It does and is equal to $-\frac 12$. Not only that, $$\lim_{x\to 0}\frac{-\sin\sqrt x}{2\sqrt x} = -\frac 12$$ as well, so not only is $\cos\sqrt x$ differentiable everywhere, its first derivative is continuous.
Checking for the second and higher derivatives leads to similar conclusion.
