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For an open set $U\subseteq \mathbb{R}^n$, given a smooth function $f:U\to \mathbb{R}$, its graph $$ G_f = \{(x,f(x)) ~|~ x\in U )\} \subseteq \mathbb{R}^{n+1} $$ is a hypersurface (codimension-1 submanifold) in $\mathbb{R}^{n+1}$. The coordinate chart on $G_f$ corresponding to the parametrization $X:U\to G_f$ given by $ X(u) = (u,f(u)) $ is called its graph coordinates. The coordinate vectors here are $$X_*\left(\partial_i\right) = \frac{\partial}{\partial x_i} + \partial_i f ~\frac{\partial }{\partial x_{n+1}} $$ and so the metric coordinates are $$ g_{ij} = \delta_{ij} + (\partial_i f)( \partial_j f) $$

The claim is that its inverse is $$ g^{ij} = \delta^{ij} - \frac{ (\partial_i f) (\partial_j f) }{ 1+\|\nabla f\|^2} $$ where $\| \cdot \|$ denotes the Euclidean norm.

I have checked this for $n=1$ but I don't know how to proceed from there.

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Maybe use the so-called Sherman–Morrison formula

https://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula

with $A={\rm Id}_{n\times n}$ (the identity matrix), and where $u=v=\nabla f$.

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  • $\begingroup$ This works. Thank you! $\endgroup$ Commented May 1 at 6:45

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