Showing cos$(x^2)$ is analytic

In the textbook that I am using I am given the following definition and theorem:

Definition: A function $f$ whose Taylor series converges to $f$ in a neighborhood of $c$ is said to be analytic at $c$.

Theorem: Let $f$ be infinitely differentiable in a neighborhood $I$ of $c$. Suppose $x \in I$ and there exists $M>0$ such that $ |f^{(m)}(t)| \leq M$ for all $m \in \Bbb{N}$ and $t \in [c,x]$ (or $[x,c] \text{ if } x<c).$ Then $\lim_{n \rightarrow \infty} R_n(x)=0$. Thus, $f$ is analytic at $c$.

Then the question I am having trouble with is showing that cos$(x^2)$ is analytic at $x=0$. What I am attempting to do is to use the formula for $R_n(x)$. Which is $\displaystyle R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$.

But trying to find a bound for all the derivatives of cos$(x^2)$ is giving me a difficult time because of having to apply the chain rule. Is there a simple technique that I am missing given this definition and theorem?