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Consider a function $A: R^n \to R^m$ (smooth) and another $v: R \to R^n$ defined as $v: R \ni u \mapsto u \cdot w $ for $w\in R^n$. Finally let $x \in R^m$.

Now define a new function $B : R \to R$ $$ u \mapsto A (v(u)) ^t x. $$

We would like to say something clever about the derivatives $$ \frac{\partial^k B}{\partial u ^k} . $$ If $n=m=1$ then by the chain rule $$ \frac{\partial^k B}{\partial u ^k} = \frac{\partial^k A}{\partial u ^k} (u w ) w^k x $$

And I'm wondering if something similar exists in general i.e. if $\tfrac{\partial^k B}{\partial u ^k}$ can be expressed as some sort of linear combination of mixed partial derivatives of $A?$ (and powers of $w$). I'm comfortable with multiindices.

What if $m=n$ or $m=1$?

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Use the multi-variable chain rule: if $f \colon \Bbb{R}^k \to \Bbb{R}^m$ and $g \colon \Bbb{R}^m \to \Bbb{R}^n$, then for $v \in \Bbb{R}^k$, then $d(g \circ f)_v$, the Jacobian matrix of $g \circ f$ at the point $v$, is given by:

$$ d(g \circ f)_v = dg_{f(v)} \circ df_v $$

In your case we can think of this as a three-step composition. In your notation from above, let's call $C \colon \Bbb{R}^m \to \Bbb{R}$ the map which takes the inner product with the fixed vector $x$. So for $y \in \Bbb{R}^m$, $C(y) = \left<y, x\right>$. Or maybe $C(y) = y^tx$ in the notation you use above. Then your function $B$ is the three-step composition:

$$ B = C \circ A \circ v $$

I'm not sure what you mean by the "partial derivatives" of $B$, though. It is just a function from $\Bbb{R}$ to $\Bbb{R}$, so there is just the one ordinary single-variable derivative $\frac{dB}{du}$. We can express this derivative as a composition using the chain rule:

$$ \frac{dB}{du} = dC_{A(v(u))} \circ dA_{v(u)} \circ dv_u $$

Unfolding notation, you can see that $dv_u$ is just the vector $w$ (written as a column vector), $dA_{v(u)}$ is the jacobian matrix of $A$ (with entries $\frac{\partial A_i}{\partial x_j}$) at the point $uw$, and $dC_{A(v(u))}$ is the vector $x$ (written as a row vector). So taking all this into account, we can re-write the chain rule above in this case as

$$ \frac{dB}{du} = \left< x, dA_{uw} w\right> = x^t dA_{uw} w $$

If you want, you can write this out with indices by doing the matrix multiplication. Let's call $y_1,\dots,y_n$ the coordinates on $\Bbb{R}^n$. Then this would be:

$$ \frac{dB}{du} = \sum_{i=1}^m \sum_{j=1}^n x_i w_j \frac{\partial A_i}{\partial y_j}(uw) $$

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