Why is this integral from Giusti's monograph valid?

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.