Given $\frac{f(x)}{g(x)} = \frac{\sin(x)}{1+\cos(x)}$, with no domain constraint. If I take the derivative of $\frac{f(x)}{g(x)}$, can the denominator be ZERO in the quotient Rule/Formula of derivatives?
If yes, what does it signifies? I had $\frac{dy}{dx} = \frac{1+\cos(x)}{(1+\cos(x))^2}$ I have a concern that if I cancel the numerator's and denominator's $(1+\cos x)$, for $x=(2n-1)\pi$, the $1+\cos(x)$ will be $0$. Hence, it is not okay to cancel. Is my understanding needs any missing piece?
Adding one more concern: What if I cancel out $(1+\cos x)$ from both the numerator and denominator, do I require to mention that $x$ cannot be equal to $(2n-1)\pi$ while talking about any order derivative in this particular case?
We should perhaps start with the observation that $$ \frac{f(x)}{g(x)} = \frac{\sin x}{1+\cos x} $$ cannot be a complete definition of a function over all real numbers. At each value of $x$ where $\cos x = -1$ (that is, whenever $x = (2n-1)\pi$ for integer $n$) the denominator is zero and the quotient $f(x)/g(x)$ is not defined.
It is not even possible to provide additional definitions of $f(x)/g(x)$ at these values of $x$ that will give us a continuous function at those points, let alone a derivative, because $$ \lim_{\large x\to\pi^-} \frac{\sin x}{1+\cos x} = +\infty $$ but $$\lim_{\large x\to\pi^+} \frac{\sin x}{1+\cos x} = -\infty. $$
That is, the function has a vertical asymptote at $x=\pi.$ There is a similar asymptote at $x = (2n-1)\pi$ for every integer $n.$ So if this function is to be defined as a differentiable function at all then there must be a domain constraint of some sort in order to eliminate the values of $x$ where $\cos x = -1.$
In some contexts it is assumed that the domain of a function described by a closed formula is all of the real numbers except the values of $x$ where the formula itself is undefined. In such a context it would be fair to present a definition of a function using only the expression $(\sin x)/(1+\cos x),$ since the domain restriction $x \neq (2n - 1)\pi$ would be considered an implicit part of the definition. In that context it probably ought to be considered legitimate to write the derivative of the function using the same implicit domain restriction, that is, to write
$$ \frac{\mathrm d}{\mathrm dx}\frac{f(x)}{g(x)} = \frac{1}{1+\cos x}, $$
where you use the fact that $1/(1 + \cos x)$ is undefined for $x = (2n - 1)\pi$ to implicitly exclude all such values of $x$ from the domain (which conveniently justifies canceling one factor of $1 + \cos x$ in order to arrive at this simplified form of the derivative).
On the other hand, there is also a tradition of giving exercises in textbooks in which a function "definition" is given in the form of a single formula and the problem is to find the values of $x$ that must be excluded from the domain. In these exercises one must not prematurely cancel factors in the numerator and denominator of a quotient. More importantly, the student is required to explicitly identify a minimal set of restrictions that must be placed on the domain in order to make the function defined by the given formula.
So while I think it would be a bit unfair of someone to present a function using an implicit definition of the function's domain and expect you to provide the derivative with an explicit definition of the derivative's domain, it might be a good idea to explicitly specify the domain of the derivative when giving your answer. The domain is, after all, something that you really ought to consider as part of the mathematics of the derivative in this exercise, and you absolutely do need to consider the domain of the derivative before you do any cancellation of factors in the numerator and denominator in order to simplify the formula.
So while I can only guess about the true intentions of whoever posed this problem (and more importantly, if this is a school exercise, the interpretation of whoever will be grading the response), my strategy would be to state the obvious explicit domain restriction of the original function as soon as I could, and to refer to that domain restriction wherever it is relevant, both to ensure a factor is not zero when canceling it and to explicitly state the domain in the final answer. That at least is a fully defensible approach.