It depends what you're graphing.
The buttons on the left of your image say "xCurve, yCurve, zCurve", and this suggests that you have a 3D curve, and you are graphing one of the coordinates ($x$) versus a time parameter, $t$.
If so, the graph of the derivative is certainly wrong. It should have the value $0$ when the abscissa (the horizontal axis value, the $t$-value) is around 17 or 53.
On the other hand, your graphs don't look like "$x$ versus $t$" graphs, they look like "$(x,y)$ versus $t$" graphs. If this is the case, then your results might well be correct (though undesirable). See below for details.
Let's start from the beginning with a nice simple notation:
Suppose $P(t)$ is a cubic Bezier, with control points $A$, $B$, $C$, $D$. Then its equation is:
$$P(t) = (1-t)^3A + 3t(1-t)^2B + 3t^2(1-t)C + t^3D \quad (0 \le t \le 1) $$
Then the derivative curve is a quadratic (degree 2) curve, and its control points are $3(B-A)$, $3(C-B)$, $3(D-C)$, so it's equation is:
$$Q(t) = 3(1-t)^2(B-A) + 6t(1-t)(C-B) + 3t^2(D-C) \quad (0 \le t \le 1) $$
All of this applies regardless of whether $A$, $B$, $C$, $D$ are $x$ values or $(x,y)$ values.
If you want to draw "$x$ versus $t$" graphs, then drawing $P$ and $Q$ together on the same graph should be straightforward.
If you want to draw "$(x,y)$ versus $t$" graphs, then putting both $P$ and $Q$ on the same graph is more problematic. Suppose the control points $A$, $B$, $C$, $D$ were a great distance from the origin, but fairly close to each other. Then $B-A$, $C-B$, $D-C$ would be small, so the $Q$ curve would be close to the origin -- far away from the $P$ curve. In your case, it looks like (roughly) $A = (0,0)$ and $B=(32,12)$, so the first control point of the derivative curve $Q$ is $3(B-A) = (96,36)$, which is off the charts. Your derivative graph is clipped, so the end-points of the curve are not visible, which makes it harder to say whether or not it's correct. At least it looks like a parabola, though, which is correct (Bezier curves of degree 2 are parabolas).
These notes might help. Section 2.5 discusses derivatives, and there's a picture showing how the derivative curve relates to the original one (for the xy-vs-t case). Section 2.12 talks about the x-vs-t type of curve (which is variously referred to as a real-valued, explicit, or non-parametric Bezier curve).
I suspect there is no natural sigmoid function that will "preserve" in some sense your original function $f$.
In all generality, you want to "squeeze" the real line ($\mathbb{R}$) into an (open) interval of the form $(x_1,x_2)$. Let's try to squeeze both the $x$-axis and the $y$-axis so that we may fit one graph into a rectangle. Let this rectangle be, for simplicity, $(-a,a)\times(-b,b)$ where $a$ and $b$ are positive.
Let $D$ be the domain of $f$ and $R$ be its codomain (for the sake of simplicity, let's think of $R$ as $\mathbb{R}$)
To squeeze the $x$-axis we really need a function $\varphi:\ (-a,a)\longrightarrow D$, so that when we graph $f\circ\varphi$ we'll need to see what that thing does only on the interval $(-a,a)$.
To squeeze also the $y$-axis, we need another function $\psi:R\longrightarrow(-b,b)$. The end result is then
\begin{equation}
\tilde{f}:=\psi\circ f\circ\varphi
\end{equation}
Its graph will be contained in the rectangle $(-a,a)\times(-b,b)$.
The tricky part, as you pointed out, is finding the "right" sigmoid function(s). One can try various cocktails. If $D=R=\mathbb{R}$ for instance, one could take
\begin{equation}
\varphi(x):=k\tan\left(\frac{\pi}{2a}x\right)\qquad\psi(x):=\frac{2b}{\pi}\arctan(x/k)
\end{equation}
to do the job. Varying the parameter $k$ allows for trying to "match" the original function (for biggish $k$). It really depends on what characteristics of $f$ you want to study.
[EDIT: I personally like viewing functions "wrapped" on a sphere- but it's only an aesthetic approach. Here i posed a question about it.]
which returns 0 if x>n; 1 if x<=n. This might help with the conditional function I'm looking for. Thank you.
Best Answer
I'll preface this by stating that I'm not entirely sure if I understand what kind of function you are trying to achieve.
Some points I noticed:
I think that the path of least resistance for you would be to code the function in C++ and output its values to a .csv file and then plot that.