Is the property "being a derivative" preserved under multiplication and composition? Since differentiation is linear, we therefore have that if $f, g: I\to \mathbb{R}$ is a derivative (where $I\subset \mathbb{R}$ is an interval), then so does their linear combination. What if we consider their multiplication and composition?
Due to the forms of the product rule of differentiation of product function and chain rule of differentiation of composition, I highly doubt their product or composition necessarily is still a derivative, but I cannot construct counterexamples.

From the product rule of differentiation or from the rule of partial integration, one obtains the following less known product rules:
\begin{equation}\int f(x)g(x)dx=f(x)G(x)-\int f'(x)G(x)dx\hspace{1.5cm}[Borman], [Goldstein], [Kowalk]\end{equation}
\begin{equation}\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx\hspace{3.9cm}[Mercer], [Reinsch]\end{equation}
$F(z)=\int f(z)dz$; $G(z)=\int g(z)dz$; $\gamma$: compositional inverse of the function $g$
From the chain rule of differentiation, the inverse rule of differentiation and the rule of partial integration, one obtains the following less known chain rules:
\begin{equation}\int f(g(x))dx=\int f(t)\gamma'(t)dt;\ t=g(x)\end{equation}
\begin{equation}\int f(g(x))dx=xf(g(x))-\int f'(t)\gamma(t)dt;\ t=g(x)\end{equation}
\begin{equation}\int f(g(x))dx=\left(\frac{d}{dx}F(g(x))\right)\int\frac{1}{g'(x)}dx-\int \left(\frac{d^{2}}{dx^{2}}F(g(x))\right)\int\frac{1}{g'(x)}dx\ dx\end{equation}
\begin{equation}\int f(g(x))dx=\frac{F(g(x))}{g'(x)}+\int F(g(x))\frac{g''(x)}{g'(x)^{2}}dx\end{equation}
This formulas show that integration rules for products and compositions of functions are possible only for some special kinds of component functions $f$ and $g$.
