Proving every derivative of $\sqrt\cos x$ is unbounded? This question is related to one I posted earlier today. I am $99$% sure the claim is true, and I can of course prove it for the first two or three derivatives, but I don't know how to jump to the infinite case. Is induction of some kind in order? I'm not sure how to proceed.
 A: Suppose $f:(a,b)\to\mathbb R$ is unbounded above and differentiable. Claim: $f': (a,b)\to \mathbb R$ is unbounded. 
Proof: Let $n\in \mathbb N$. Our goal is to find $c\in (a,b)$ such that $ f'(c) >n$. Let $z=\frac{b-a}{2}$. Since $f$ is unbounded above, there exists $x\in (a,b)$ such that $f(x) > (b-a)n+f(z)$. Suppose $x>z$; if $x<z$, then the proof will work similarly (although $f'$ will be unbounded below), and if $x=z$, then we can choose some other $x$ that satisfies the inequality. Then by the mean value theorem, there exists $c\in (z,b)$ such that
$$f'(c) = \frac{f(x)-f(z)}{x-z} \geq \frac{f(x)-f(z)}{b-a} > \frac{1}{b-a}((b-a)n+f(z)-f(z))=n$$
Since this works for any $n$, $f'$ is unbounded. A similar claim holds if $f$ is unbounded below. 
So, since $\sqrt{\cos x}$ is infinitely differentiable and since its first derivative is unbounded, it follows (using induction) that each higher derivative is unbounded. 
A: $f(x) = \sqrt{\cos x}\implies |f'(x)| = \dfrac{|\sin x|}{2\sqrt{\cos x}}$. We have: $|f'(x)| \to +\infty$ when $x \to \frac{\pi}{2}^{-}$, this is enough to show $f'(x)$ is not bounded. 
A: let $c=\sqrt{\cos 2x}$, $s=\sqrt{\sin 2x}$, $t=\sqrt{\tan 2x}$, and let $D$ be the differentiation operator w.r.t. $x$, so:
$$
\begin{align}
Dc &= -st \\
Ds &= \frac{c}{t} \\
Dt &= t^3+\frac1t
\end{align} 
$$
a little algebra, together with Leibniz' rule for differentiating a product, shows that as $x \to \frac{\pi}4$ and $t \to \infty$ the differential coefficient $D^n c$ is dominated by the term $-st^{2n-1}$
