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In this link, it says that the gradient is as follow
The Gradient of the TV norm is $$ \mathrm{Grad}J(f)=\mathrm{div}\left(\frac{\nabla f}{\lVert\nabla f\rVert}\right). $$
From this other link, it also mentions that the derivative is as follow
Formally, we can write $\partial J(u)=-\mathrm{div}\left(\frac{\nabla u}{\lvert \nabla u\rvert}\right)$
I know how to calculate divergence, but I don't understand how the gradient of total variation is related to divergence.
I am by no means an expert on total variation, however I think you should check out this Wikipedia page. It doesn't directly answer your question, but I believe the lemma below illustrates the relationship between total variation and divergence.
There, it gives a lemma that follows from the Gauss-Ostrogradsky theorem and provides a proof for it,
$\int_{\Omega}f\,div(\phi) = -\int_{\Omega} \nabla f \cdot\phi$.
I think for some intution about divergence, it might be helpful to read Wikipedia's explanation.
To obtain the Gradient of the TV norm, you should refer to the calculus of variations. By examining the TV minimization with Euler-Lagrange equation, e.g,, Eq. (2.5a) in [1], you would see the answer.
[1] Nonlinear total variation based noise removal algorithms, 1992.
