Derivative of an integral of differential form

I have some smooth function $g(x) \colon \mathbb{R}^{n}_{+} \to \mathbb{R}_+$ such that $G_{t} = \{ x \in \mathbb{R}^n_+ \mid g(x) \leqslant t \}$ is compact. I consider a function $$f(t) = \int\limits_{G_t}a(x)dx_1 \wedge ... \wedge dx_n$$ I want to find its derivative.

In this article http://amath.colorado.edu/pub/wavelets/papers/BEYLKI-1984.pdf author uses the represenation of the form $dx_1 \wedge ... \wedge dx_n = dg(x) \wedge \Omega$ to reduce an integral of the form $dx$ to an iterated integral. Is it possible to do something similar here? I think the answer is $$f'(t) = \int\limits_{ \{ x\mid g(x)=t \}} a(x) \Omega$$

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1. Let's make a substitution: $$(x_1,x_2,...,x_n) \mapsto (g(x),x_2,x_3,...,x_n)$$ with Jacobian $J = \frac{1}{\partial_{1} g(x)}$. Then we can rewrite $f(t)$ as $$f(t) = \int\limits_{0}^{t} \frac{a(x_1(\tau;y);y) d\tau}{\partial_{1}g(x_1(\tau;y);y)} \int\limits_{\mathbb{R}^{n-1}_+}dy$$ where $y = (y_1,...,y_{n-1})$ and $x_1(\tau;y)$ is defined by $g(x_1(\tau;y);y) = \tau$. It is easy to differentiate $f(t)$ now: $$f'(t) = \int\limits_{\mathbb{R}^{n-1}_+} \frac{a(x_1(t;y);y)dy}{\partial_1 g(x_1(t;y);y)}$$
2. We can consider $M_t = \{ x \in \mathbb{R}^n_+ \mid g(x) = t \}$ as a smooth manifold with the atlas $\{ (M_t,\varphi) \}$ such that $\varphi \colon M_t \to \mathbb{R}^{n-1}_+$ and $\varphi(x_1,...,x_n) = (x_2,...,x_n)$. Then a form $\omega$ defined by $$\omega = \frac{a(x_1(t;y);y)}{\partial_1 g(x_1(t;y);y)} dy_1 \wedge ... \wedge dy_{n-1}$$ is a pullback via $(\varphi^{-1})^{*}$ of a form $\alpha$ defined by $$\alpha = \frac{a(x)}{\partial_1 g(x)} dx_2 \wedge ... \wedge dx_n$$ Hence $$f'(t) = \int\limits_{M_t} \alpha$$
3. Since $dg \wedge \omega_1 = dg \wedge \omega_2$ implies $$\int\limits_{M_t} \omega_1 = \int\limits_{M_t} \omega_2$$ and $dg \wedge \alpha = a(x) dx_1 \wedge ... \wedge dx_n$ than for any form $\beta$ such that $dg \wedge \beta = a(x)dx$ we have $$f'(t) = \int\limits_{M_t} \beta$$ For example, we can choose $\beta$ defined by $$\sum\limits_{k=1}^{n} (-1)^{k-1} \frac{a(x)\partial_k g(x)}{| \nabla g(x) |^2} dx_1 \wedge ... \wedge \overline {dx_k} \wedge ... \wedge dx_n$$