Trouble finding kernel of a linear transformation $T_g: C^1[0, 1] \rightarrow C[0, 1]$ 
Let $T_g: C^1[0, 1] \rightarrow C[0, 1]$ be the linear transformation
$$T_g(f) = (fg)'$$
Find the null space of $T_g$.

I'm having trouble formulating a correct answer to this problem. It is straightforward to note that $T_g(f) = 0 \iff (fg)' = 0$ and by the rules of differentiation this is true iff $f'g + g'f = 0$. Now it might be observed that
\begin{align*}
    &(f'g)(x) + (g'f)(x) &= 0 \\
    \iff&f'(x)g(x) + g'(x)f(x) &= 0 \\
    \iff&\int f'(x)g(x) ~dx + \int g'(x)f(x) ~dx &= \int 0 ~dx \\
    \iff &f(x)g(x) - \int g'(x)f(x) ~dx + \int g'(x)f(x) ~dx &= C \\
    \iff&f(x)g(x) &= C
\end{align*}
where $C \in \mathbb{R}$ is an arbitrary constant. (Notice that from the third to the fourth line we used integration by parts.)
I : I am unsure of whether this result is correct. I'm quite new to linear algebra and I'm not confident on how valid my integration approach is.
II : Even if it is correct, the result $fg = c$    for $c$ a constant depends both on $f$ and on $g$. However, the kernel is dependent on the domain of $T_g$, whose elements are mapped onto the vector $f$. In other words, it seems dubious to define the kernel as
\begin{equation*}
    kern(T_g) = \Big\{f \in C^1[0, 1] : f = c_1, g = c_2 \mid c_i \in \mathbb{R}\Big\}
\end{equation*}
since $g$ is not a variable.
I'm sure I'm making this more complicated than it needs to be. Any help is appreciated.
 A: 
I : I am unsure of whether this result is correct. I'm quite new to linear algebra and I'm not confident on how valid my integration approach is.

You have the right idea, but I think you're thinking a little too hard as well. My thought process would have been like:

*

*"We need $(fg)' = 0$."

*"Well, derivatives of constants are zero."

*"Hence, $f$ needs to be such that $fg$ is a constant."

No real need for detailed calculations to make that realization.

II : Even if it is correct, the result $fg = c$    for $c$ a constant depends both on $f$ and on $g$. However, the kernel is dependent on the domain of $T$, whose elements are mapped onto the vector $f$. In other words, it seems dubious to define the kernel as
\begin{equation*}
    kern(T) = \Big\{f \in C^1[0, 1] : f = c_1, g = c_2 \mid c_i \in \mathbb{R}\Big\}
 \end{equation*}
since $g$ is not a variable.

While the result depends, in some sense, on $g$, I think the original question is ill-framed. Notice that the domain of $T$ is $C^1[0,1]$, not $C^1[0,1] \times C^1[0,1]$. This is meant to indicate that $f$ is your variable, not $f$ and $g$. This is, in part, why the notation $T_g$ was used, but that should have been maintained throughout the problem statement.
In other words, treat $g$ as a fixed and given function; what must $f$ be so that $(fg)' = 0$? That is, no matter how $g$ is defined (so long as it is differentiable, obviously), how can you define $f$ so that $fg=c$ for a constant $c$?
