Suppose we have polynomials $f$ and $g$ over a field $F$ and for some $a\in F$ we know $(x-a)^m$ divides $f\circ g$, i.e. $f\circ g$ has a zero of order $m$ at $a$. Then we have that $(x-a)^{m-n}$ divides $(f\circ g)^{(n)}$, the $n$'th derivative of $f\circ g$. Now suppose we know that $g'$ is relatively prime to $x-a$. By the chain rule $(x-a)^{m-1}$ divides $f'\circ g \cdot g'$ and so must divide $f'\circ g$. Inductively, one can show then that $(x-a)^{m-n}$ divides $f^{(n)}\circ g$. 


My question is: is there a reasonably simple condition under which this works when $f$ is multivariate? Specifically, if $f(x\_1,\ldots,x\_l)\in F[x\_1,\ldots,x\_l]$ and $g\_1,\ldots,g\_l$ are univariate then what is the order of the zero of $f^{(n\_1,\ldots,n\_l)}(g\_1,\ldots,g\_l)$ at $a$ given the order of the zero of $f(g\_1,\ldots,g\_l)$ at $a$? Here $f^{(n\_1,\ldots,n\_l)}$ is the result of applying the partial derivatives $\frac{d^{n\_i}}{dx\_i^{n\_i}}$ to $f$.



Zeros of compositions of polynomials and derivatives
This doesn't work in general. It only works if we differentiate $f$ along the direction of $g$, that is, apply the multivariate chain rule to deduce results like


$\sum\_{i=1}^l g\_i' \frac{df}{dx\_i}=0$


$\sum\_{i=1}^l\sum\_{j=1}^l g\_i' g\_j' \frac{d^2f}{dx\_idx\_j} +\sum\_{i=1}^l g\_i'' \frac{df}{dx\_i}$


We cannot reason about the original derivatives.


What's going on is that the image of $(g\_1,...,g\_l)$ is a curve in space. We are evaluating the function on the curve, and computing the derivatives of the function on the curve . But this only allows us to compute the derivatives along the curve. We have no way to say anything about the derivatives in another direction. It is easy to check that $f(g\_1,\dots g\_l)$ could be completely zero, vanishing to infinite order, while many of the derivatives of $f$ do not even vanish to first order!


