Is the product of two derivative functions still a derivative function? Is the product of two derivative functions still a derivative function? I.e., given two differentiable functions $f$ and $g$, is there always a differentiable function $k$ with $f'g'= k'$  ?
 A: The answer in general is no, here is a counterexample, where we use the same function $f(x) = g(x) = x^2 \sin \frac{1}{x^2}$ for $x \ne 0$, and $f(0) = g(0) = 0$. This function is differentiable on the whole real line with $f'(x) = 2x \sin \frac{1}{x^2} - \frac{2}{x} \cos \frac{1}{x^2}$ for $x \ne 0$, and $f'(0)=0$. However, the product $f'(x)g'(x) = f'(x)^2$ satisfies $\int_0^1 f'(x)^2 \, dx = +\infty$, so if there was a function $k$ with $k'(x) = f'(x)^2$, then by the Fundamental Theorem of Calculus (using the fact that $k'$ is continuous everywhere except at $0$), we would get $k(1) - k(0) = +\infty$, so there can not be such a function $k$.
The integral divergence is not trivial to see, but it follows from the facts that $f'(x)^2 \ge 0$, that the "envelope" of $f'(x)^2$ grows like $\frac{4}{x^2}$, with $\int_0^1 \frac{4}{x^2} \, dx = +\infty$, and that $f$ oscillates somewhat regularly.

ADDENDUM: Thanks to Dave L. Renfro for pointing out the survey article Some aspects of products of derivatives by Andrew M. Bruckner, Jan Mařík, and Clifford E. Weil [American Mathematical Monthly 99 #2 (February 1992), 134-145]. It summarizes some related research motivated by this question. In the introduction they point to a counterexample given in the paper Some properties of derivative functions by Witold Wilkosz [Fundamenta Mathematicae, vol. 2(1), (1921), 145-154]. Witosz shows that there exists a function $f$ such that $f'(x) = \cos \frac1x$ for $x \ne 0$, and $f'(0)=0$, but that there does not exist a function $k$ such that $k'(x) = \cos^2 \frac1x$ for $x \ne 0$ and $k'(0)=0$. This example also shows that counterexamples with bounded derivatives exist.
In order to see why $\cos^2 \frac1x$ is not a derivative, here is a slightly simpler argument stolen from the article When Is the Product of Two Derivatives a Derivative? by Michael W. Botsko [Mathematics Magazine, 
vol. 65(3), (1992), 186-187]. Let $F(x) = -x^2 \sin \frac1x$ for $x \ne 0$, and $F(0)=0$. Then $F'(x) = -2x \sin \frac1x + \cos \frac1x$ for $x \ne 0$ and $F'(0)=0$. The function $h(x) = x \mapsto -2x \sin \frac1x$ is continuous (with $h(0)=0$), so it is a derivative by the Fundamental Theorem of Calculus. This shows that there exists a differentiable function $f$ with $f'(x) = \cos \frac1x$ for $x \ne 0$, and $f'(0)=0$. A similar argument with $\cos$ swapped with $\sin$ shows that there exists a differentiable function $g$ with $g'(x) = \sin \frac1x$ and $g'(0)=0$. If we assume that there exist differentiable functions $k$ and $l$ with $k'(x) = f'(x)^2$ and $l'(x) = g'(x)^2$, then $k'(x) + l'(x) = \cos^2 \frac1x + \sin^2 \frac1x = 1$ for $x \ne 0$, and $k'(0)+l'(0)=0$. This implies that the function $m(x) = k(x) + l(x)$ is differentiable with $m'(x) = 1$ for $x \ne 0$, and $m'(0)=0$. However, it is easy to see that no such function exists.
Strictly speaking, this argument only shows that one of $f'(x)^2$ and $g'(x)^2$ is not a derivative, but it is not too hard to show that neither of them is (since they are almost the same function.)
