In the proof of Lemma 5.8 of Giusti's monograph "Minimal Surfaces and Functions of Bounded Variation", one considers a function $f \in C^1(B_R)$ and defines $f_t(x) = f(tx/|x|)$ at least if $x \in B_t$, the ball of radius $t$ centered at $0$. Giusti claims without proof that $$\int_{B_t} |(df_t)_x| ~dx = \frac{t}{n - 1} \int_{\partial B_t} |(df)_x|\sqrt{1 - \frac{\langle x, (df)_x\rangle^2}{|x|^2 |(df)_x|^2}} ~dS(x).$$ Here of course $S$ is the spherical measure and $n$ is the dimension of the ambient space. I think this is supposed to be "obvious" but I don't see it.
Here's something I've tried. By the chain rule, $$(df_t)_x = (df)_{tx/|x|} \circ (A_t)_x$$ where $A_t$ is the derivative of $x \mapsto tx/|x|$. At least if $n = 2$ (but probably in general), $A_t = tR$ where $R$ is the standard symplectic matrix $R = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$. In particular, $R$ is unitary, so $$|(df_t)_x| = t|(df)_{tx/|x|}|$$ and $|(df)_{tx/|x|}|$ does not depend on $t$. This suggests that we should integrate in polar coordinates $$\int_{B_t} |(df_t)_x| ~dx = t\int_0^t \int_{\partial B_s} |(df)_{tx/|x|}| ~dS(x) ~ds$$ and since the integrand of $dS(x)$ does not depend on $s$ it is tempting to apply the rescaling $\partial B_s \to \partial B_t$ and then use Fubini's theorem to dispose of the integral $ds$. This doesn't seem to help though because we need terms of the form $(df)_x$ to appear when we perform this rescaling, but the rescaling does not depend on $f$, so cannot create terms of the form $(df)_x$.
EDIT: I missed a square
Just for posterity, here's what's going on. The integral on the right-hand side simplifies to $\int_{\partial B_t} |\partial_\Theta f|$ where $\partial_\Theta$ denotes the part of the derivative which is tangential to $\partial B_t$. In polar coordinates, the left-hand side is $$\int_{B_t} |df_t| = \int_0^t \int_{\partial B_s} |df_t| ~dS_s(\Theta) ~ds = \int_0^t \int_{\partial B_t} |df_t| \frac{dS_s(\Theta)}{dS_t(\Theta)} ~dS_t(\Theta) ~ds$$ and since $\partial_r f_t = 0$ we get on $\partial B_t$ that $|df_t| = |\partial_\Theta f|$. Also $dS_s(\Theta)/dS_t(\Theta) = (s/t)^{n - 1}$, and by Fubini we can break up the left-hand side as $$\int_{B_t} |df_t| = \left[\int_0^t (s/t)^{n - 1} ~ds\right]\left[\int_{\partial B_t} |\partial_\Theta f|\right]$$ as desired.