Is it possible to write sin x as a result of its derivative?

Question

Basically, I'd like to model sin x, but make it's derivative tend towards 0, so as x increases, it becomes a constant y = 0. The function begins like a typical sin x function, but slowly the fluctuation decreases until it isn't there anymore. If this works as I'm trying to have it work, I think some constant between 0 and 1 multiplying that derivative could also re-establish the normal sin x function.

I've been messing around with desmos graphing calculator, somehow trying to make sin x a result of some equation containing it's derivative, but I haven't been able to make much progress. I have taken classes in differential and integral calculus and linear algebra.

Edit: Sorry for the lack of rigour, I'm struggling to formulate the question properly

Edit2: User @ElliotG has provided me with the exact equation I'm looking to obtain, or atleast one that perfectly describes the idea of what the equation I'm looking for looks like: $\frac{sin x}{1 + x^2}$. The way I'd describe this function is that it is like sin x, but having its derivative constantly decreasing until it reaches 0. What I'd be interested in is: could there have been a way of obtaining a similar equation to $\frac{sin x}{1 + x^2}$ having in mind that what we want to do is have sin x as if its derivative was tending to 0? So all we start by knowing is how we want the function to behave without knowing what it looks like.

Answer 1

Let $$f(x) = e^{-\alpha x} \sin x \quad \alpha <<1$$

Here is a graph of this function, and also another nice choice that was suggested $g(x)=\frac{\sin x}{1+\alpha^2 x^2}$, both with $\alpha = \frac{1}{10}$.

The derivatives are:

$$f^\prime(x)=e^{-\alpha x} [\cos x - \alpha \sin x] \textrm { and } g^\prime(x)=\frac{\cos x}{1+\alpha^2 x^2} -\frac{2x \sin x}{\alpha(1+\alpha^2 x^2)^2}$$

Here is a picture of the derivatives $f^\prime(x)$ and $g^\prime(x)$

$f$ and $f^\prime$ are in blue and $g$ and $g^\prime$ are in yellow.

Answer 2

Any function of the form $$y(t)=Ae^{-\gamma t}\sin(\omega t+\phi)$$ works. For example, substituting $\gamma = 0.2$, $A=1$, $\omega = 1$ and $\phi = 0$, we get the following curve:

which satisfies your constraints.

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Your question also pertained to why such a function works, and the reason is that this function was developed in physics as a model for the damped Harmonic Oscillator. This was initially intended as a solution to the following form of Newton's second law: $$m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+kx=0$$ and subsequently introducing an additional damping force. The equation is a method to model oscillators like pendulums in realistic scenarios (where constraints like air resistance prevent the oscillator from continuing indefinitely).