Can I simply replace the first and second derivatives with the gradient and laplacian in the curvature of the graph of a function to generalize to a multivariable curvature functional of the graph of a function?
The curvature of the graph of a function is given by the following.
$$\mathscr{k}_f(x) = \frac{f^{\prime\prime}(x)}{(1 + [f^{\prime}(x)]^2)^{\frac{3}{2}}}$$
Assuming a second-differentiable scalar function, is the following a valid generalization of this type of curvature to the multivariable case?
$$\mathscr{k}_f(\vec{x}) = \frac{\nabla_{\vec{x}}^2 f(\vec{x})}{(1 + [\nabla_{\vec{x}} f(\vec{x})]^2)^{\frac{3}{2}}}$$
Edit
Matthew Pilling convinced me that taking the square of a gradient is not directly possible. I was implicitly assuming I could collapse the vectors by summing the components, however it would be technically correct to make this explicit using either dot products or the trace of the diagonal of the resulting vectors.
$$\mathscr{k}_f(\vec{x}) = \frac{\nabla_{\vec{x}}^2 f(\vec{x}) \cdot \vec{1}}{(1 + [\nabla_{\vec{x}} f(\vec{x})\cdot \vec{1}]^2)^{\frac{3}{2}}} = \frac{\text{tr}[\text{diag}[\nabla_{\vec{x}}^2 f(\vec{x})]]}{(1 + [\text{tr}[\text{diag}[\nabla_{\vec{x}} f(\vec{x})]]]^2)^{\frac{3}{2}}}$$
This is with the diagonal of a vector $\text{Diag}(\vec{x})$ being the construction of a diagonal matrix whose diagonal entries are the corresponding components of the vector $\vec{x}$.
Beyond this correction on vector operations, I'm still interested in whether the corrected equation captures the same idea of curvature as the original equation.