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Let $a \in \mathbb{R}^{n}$ be fixed. The directional derivative of a smooth $f:\mathbb{R}^{n}\to \mathbb{R}$ is $\nabla_{a}f(x):=a\cdot \nabla f(x)$. One can easily obtain a formula for the second directional derivative $\nabla_{a}^{2}f:=\nabla_{a}(\nabla_{a}f)$ as a quadratic form: $$\nabla_{a}^{2}f(x)=a^{T} H_{f}(x) a,$$ where $H_{f}:=(\partial_{x_{i}x_{j}}f)$ is the Hessian matrix of $f$.

For this, see the elementary proof here http://mathonline.wikidot.com/higher-order-directional-derivatives also this answer here Second directional derivative and Hessian matrix

What is the formula for the repeated directional gradient $\nabla^{m}_{a} f:=\underbrace{\nabla_{a}(\nabla_{a}(\cdots \nabla_{a}f))}_{\text{$\nabla_{a}$ taken $m$ consecutive times}}$

Also, what is the formula for $\nabla_{a}^{m} (fg)$, the repeated directional derivative of the product of two smooth functions $f,g$ ?

A reference would be very useful.

Medo
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    Start by writing out the $m$th derivative of $fg$ in the one-dimensional case. You'll need to generalize that. Notice that you have a bilinear form, rather than a quadratic form, if you do $\nabla_a\nabla_b f$. You can generalize that to an tensor of rank $m$ easily enough. None of this is pretty, but conceptually it's clear. – Ted Shifrin Nov 14 '23 at 18:25
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    Dieudonné and Lang discuss the rigorous definitions of the higher multivariable derivatives in their rigorous analysis texts. I'm sure plenty of other texts do it well, too. – Ted Shifrin Nov 14 '23 at 18:31

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