While studying Itô calculus, I came across the following question. Suppose we have the canonical $d$-dimensional Wiener process $(B_t)_{t \geq 0}$, then for sufficiently smooth $h = h(x,y)$, if we put $Y_t := h(t,B_t)$, the Itô formula $$Y_t - Y_0 = \int_0^t \partial_xh(t,B_t) + \frac{1}{2} \Delta_y^2h(t,B_t) \, dt + \int_0^t \nabla _yh(t,B_t) \cdot dB_t$$ holds. We can write this in formal "differential" notation: $$dY_t = \left (\partial_x h(t,B_t) + \frac{1}{2} \Delta_y^2h(t,B_t) \right) dt + \nabla _yh(t,B_t) \cdot dB_t .$$ Somewhere in my book the process $Z_t := \exp(-\int_0^tf(B_s) \, ds)$ appeared, where $f$ is some smooth function. The author claims that $dZ_t = -fZ_t \, dt$, but meant in the sense of Itô's formula above. This confused me since I was not able to write $Z_t$ as some function of $B_t$ and apply Itô's formula (it is rather a functional of $(B_t)_{t\geq 0}$).
However, $-fZ_t $ is of course the classical derivative of the differentiable function $t\mapsto Z_t$. Is it true that these two notions coincide in this case and if yes where can I find a proof of this fact?
Remember that Ito's formula is really only used to handle the fact that $t \rightarrow B_t$ is not differentiable, or even finite variation. Here, the function we're differentiating is $Z_t := \exp(-F_t)$ where $F_t := \int_0^t f(B_s)ds$ and $t \rightarrow F_t$ is differentiable even though $B_t$ is not. It's similar to the fact that if $g: \mathbb{R} \rightarrow \mathbb{R}$ is any integrable function then $F(t):=\int_0^t f(g(s))ds$ is differentiable and $F'(t) = f(g(t))$ no matter what $g$ is, even if $g$ is not differentiable or even continuous! Here we just have that $g(t) = B_t$, so the same reasoning applies. Then $dZ_t = -f(B_t)Z_t dt$ just follows from the chain rule.