Stability estimates on quotients of the form $ \frac{\prod_{j=1}^n a_j}{\prod_{j=1}^n b_j} $ Suppose that $a_j,b_j \in \mathbb C$ are complex numbers, $j=1,\dots,n$, with the property that $|a_j|,|b_j| \geq c > d >0$ where $c,d$ are positive real numbers. I'm interested in the stability of the product
$$
\frac{\prod_{j=1}^n a_j}{\prod_{j=1}^n b_j}
$$
under noise in the following sense: Let
$$
\tilde{a}_j = a_j + \delta_j
$$
and
$$
\tilde{b}_j = a_j + \gamma_j
$$
where $\delta$ and $\gamma$ is $\ell^2$ noise such that $\| \delta \|_{\ell^2},\| \gamma \|_{\ell^2} \leq d$. The assumption on the $\ell^2$ norm to be smaller than $d$ implies that the quotient
$$
\frac{\prod_{j=1}^n \tilde a_j}{\prod_{j=1}^n \tilde b_j}
$$
is well-defined. The difference
$$
\left | \frac{\prod_{j=1}^n \tilde a_j}{\prod_{j=1}^n \tilde b_j} - \frac{\prod_{j=1}^n a_j}{\prod_{j=1}^n b_j} \right |
$$
tends to zero as the noise tends to zero (w.r.t the $\ell^2$-norm). I'm searching for a (possibly sharp) estimate on this difference in terms of some norm of the noise and $n$. Is there a way to get such inequalities? This question seems to be quite elementary so maybe there exists some literature on this?
 A: Perhaps the simplest approach is to consider the problem as the calculation of a functional derivative. Let's put
$$
(a_1, \ldots, a_n, b_1, \ldots, b_n)\triangleq (\mathbf{a}, \mathbf{b})\in\Bbb C^{2n}
$$
and, in an analogous fashion,
$$
(\delta_1, \ldots, \delta_n, \gamma_1, \ldots, \gamma_n)\triangleq (\boldsymbol{\delta}, \boldsymbol{\gamma})\in\Bbb C^{2n}
$$
Then, defining the complex valued functional
$$
\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})\big]= \frac{\prod_{j=1}^n a_j}{\prod_{j=1}^n b_j}
$$
it is natural to write
$$
\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]= \frac{\prod_{j=1}^n (a_j+\epsilon\delta_j)}{\prod_{j=1}^n (b_j+\epsilon \gamma_j)}\quad \forall \epsilon >0
$$
and calculating the functional derivative as usual
$$
{\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma}) \triangleq \left.\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]\right|_{\epsilon=0}
$$
This procedure occasionally shows up in problems of sensitivity analysis in signal processing. This functional derivative is linear respect to the variation $(\boldsymbol{\delta}, \boldsymbol{\gamma})$: if it is a Fréchet derivative, it is possibly relatively easy to obtain a nice estimate in the form
$$
\left|\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]-\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})\big]\right|\le c\left|{\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma})\right|\label{1}\tag{☆}
$$
with a sharp constant $c$. However, even if it is a Gâteaux derivative (i.e. it depends on the structure of the variation $(\boldsymbol{\delta}, \boldsymbol{\gamma})$) you may able to obtain a sharp estimate.
Addendum. Without going in the details and assuming the (plausible) non degeneracy hypothesis $\left|{\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma})\right| >0$ when $(\boldsymbol{\delta}, \boldsymbol{\gamma})\neq 0$, further fairly elementary steps for possibly arriving at \eqref{1} are the following ones:
$$
\begin{split}
\left|\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]-\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})\big]\right| &= \int\limits_0^1\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]\mathrm{d}\epsilon \\
& \le \int\limits_0^1\left| \frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big] \right| \mathrm{d}\epsilon \\
& = \left| {\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma})\right|\int\limits_0^1\left| \frac{\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]}{\left.\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]\right|_{\epsilon=0}} \right| \mathrm{d}\epsilon \\
& \le c\left|{\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma})\right|
\end{split}
$$
and therefore the relation
$$
\int\limits_0^1\left| \frac{\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]}{\left.\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]\right|_{\epsilon=0}} \right| \mathrm{d}\epsilon\le c \label{2}\tag{☆☆}
$$
may possibly be a first guess for the sought for constant. Note that the integral on the right side of \eqref{2} can be difficult to estimate: however, despite requiring possibly tedious calculations, you can made explicit its form.
