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Let us have two function $n(\alpha)$ and $s(\alpha)$ and a set of two implicit equations $F_1(\alpha, n(\alpha), s(\alpha))=0$ and $F_2(\alpha, n(\alpha), s(\alpha))=0$. In this paper I've been reading they state that one can use the implicit function theorem to this set of implicit equations and gain $ \frac{\partial n}{\partial \alpha}=-\frac{det A}{det B} $ where $A=\begin{bmatrix} \frac{\partial F_1}{\partial s} & \frac{\partial F_1}{\partial \alpha} \\ \frac{\partial F_2}{\partial s} & \frac{\partial F_2}{\partial \alpha} \end{bmatrix} $ $B = \begin{bmatrix} \frac{\partial F_1}{\partial s} & \frac{\partial F_1}{ \partial n} \\ \frac{\partial F_2}{\partial s} & \frac{\partial F_2}{\partial n} \end{bmatrix}$

I cannot comprehend how they gained that partial derivative of n. Would be grateful for any advice.

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Let's replace $F_1$ and $F_2$ with $F$ and $G$. Differentiating each implicit equation, and using $F_1(u, v, w)$ to denote the derivative of $F$ with respect to its first argument, etc.) we get \begin{align} \newcommand{\al}{\alpha} 0 &= F_1(\al, n(\al), s(\al)) + F_2(\al, n(\al), s(\al))n'(\al) + F_3(\al, n(\al), s(\al))s'(\al)\\ 0 &= G_1(\al, n(\al), s(\al)) + G_2(\al, n(\al), s(\al))n'(\al) + G_3(\al, n(\al), s(\al))s'(\al) \end{align} where I've used $n'(\al)$ to denote the thing you've called $$ \frac{\partial n}{\partial \alpha}, $$ but which should probably be $$ \frac{dn}{d\al} $$ because $n$ appears to be a function of a single variable. Rewriting those two equations a little, and getting rid of all the arguments, we have

\begin{align} 0 &= F_1 + F_2 n' + F_3 s'\\ 0 &= G_1 + G_2 n' + G_3 s' \end{align} which we can rewrite in matrix form as $$ \pmatrix{0\\0} = \pmatrix{ F_1 & F_2 & F_3 \\ G_1 & G_2 & G_3 } \pmatrix{1\\n'\\s'} $$ Adding a row to make this $3 \times 3$, we get $$ \pmatrix{0\\0\\1} = \pmatrix{ F_1 & F_2 & F_3 \\ G_1 & G_2 & G_3 \\ 1 & 0 & 0 } \pmatrix{1\\n'\\s'} $$ and calling the $3 \times 3$ matrix $M$, we can solve $$ M^{-1} \pmatrix{0\\0\\1} = \pmatrix{1\\n'\\s'} $$ And we see that all we really need is the second entry of $$ M^{-1}\pmatrix{0\\0\\1}. $$ The authors have applied Cramer's rule to deduce that it is the ratio of determinants that they wrote down (I assume).

Reasonable questions are: "How did you know that adding [1 0 0] gave you an invertible matrix?" and "How did you decided to do that, anyhow?"

For the first: I didn't know that. But the authors don't seem to worry that $\det B$ might be zero, so I figure we're just doing some handwaving in some kind of generic case for the time being.

For the second: A ratio of determinants reminded me of Cramer's rule, so having written things in $3 \times 2$ matrix form, I just wanted to add the simplest possible thing to get a $3 \times 3$ form.

I'll be honest...I didn't actually go through Cramer's rule and check that this all works out. But I'll bet it does...

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  • $\begingroup$ Thank you very much, that´s exactly what I needed :) ..and yes there should be $dn/d\alpha$ $\endgroup$ Commented Jan 17, 2021 at 23:49

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