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Willie Wong's idea solves the problem. Here's a sketch of the argument.

Edit: there is a mistake (see bold below), so I'm no longer sure if this works.

Define $G(f)$ to be $\exp(-\int_0^1 dx/f(x))$ if $f$ is non-zero on $[0,1]$, and let $G(f) = 0$ otherwise. Let $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0, \infty))$. Without loss of generality, we'll show that all derivatives of $G(f)$ with respect to $t$ at $t=0$ exist. If $f(x,0)$ is nonzero on $[0,1]$, then this is also true for $t$ in a neighbourhood of $0$, and $G(f)$ is smooth in this neighbourhood by differentiating under the integral.

If $f(x,0)$ has a zero in $[0,1]$, then $G(f)$ is zero when $t=0$, and one can show that there exists $c > 0$ such that $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \frac{c}{|t|} \tag{*} $$ for all $t \in [-1,1]$ such that $f(x,t)$ is nonzero for all $x \in [0,1]$. Therefore, $$ \exp \left( -\int_0^1 \frac{1}{f(x,t)} dx \right) \leq \exp \left( -\frac{c}{|t|} \right) $$ for such $t$. Therefore, $$ G(f) \leq \exp \left( -\frac{c}{|t|} \right) $$ for all nonzero $t \in [-1,1]$, since either $G(f) = 0$, or $G(f)$ is as in the previous equation. It follows that the first derivative of $G(f)$ exists and is zero at $t = 0$.  Can the argument be adapted to handle the higher derivatives?

The meat is in inequality (*), which is proved using a Taylor approximation to $f$ in the $t$ variable.

Clarification on equation (*):

Let $g(t) = \min_{x\in [0,1]} f(x,t)$ (recall that $f$ is by assumption nonnegative). Since $[0,1]$ is compact and $f$ is continuous, we have that $g$ is continuous. Therefore $g^{-1}(0)$ is a closed set. The argument above the cut shows differentiability of $G(f)(t)$ away from the boundary $\partial g^{-1}(0)$.

Suppose $t = 0$ is at the boundary, such that $g(t) > 0$ in some small interval $(0,t_0)$. Now let $x_0 \in [0,1]$ be such that $f(x_0,0) = 0$. Then by smoothness there exists $\epsilon > 0$ and $C > 0$ such that

$$ f < C (t^2 + (x - x_0)^2)$$

for every $t \in [0,\min(t_0,\epsilon)]$ and $x\in (x_0 - \epsilon, x_0+\epsilon)\cap [0,1]$. The squares comes from the fact that $f$ is assumed to be a non-negative smooth function, so $\partial_xf (x_0,0) = \partial_t f(x_0,0) = 0$. So integrating we get

$$ \int_0^1 \frac{1}{f} dx \geq \int_{\max(x_0 - t,0)}^{\min(x_0 + t,1)} \frac{1}{f} dx \gtrsim \frac{t}{t^2} = \frac{1}{t} $$

as claimed.

That $f$ is non-negative is crucial here: the argument given above does not work if $f$ can be real valued, and this can be seen explicitly for the example $f(x,t) = x-t$. However, in such a situation we should demand instead of $1/f$ in the integrand, that we use $1 / f^2$.

Here is the proof that either of the definitions of $G$ in the question satisfy a weaker version of the smoothness condition. I hope that the same ideas can be used to prove that the second definition is in fact smooth. So we'll discuss that case.

Define $G(f)$ to be $\exp(-\int_0^1 dx/f(x))$ if $f$ is non-zero on $[0,1]$, and let $G(f) = 0$ otherwise. Let $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0, \infty))$. We'll show that the first derivative of $G(f)$ with respect to $t$ exists for all $t$. Without loss of generality, we'll do this at $t=0$. If $f(x,0)$ is nonzero on $[0,1]$, then this is also true for $t$ in a neighbourhood of $0$, and $G(f)$ is smooth in this neighbourhood by differentiating under the integral.

If $f(x,0)$ has a zero in $[0,1]$, then $G(f)$ is zero when $t=0$, and we'll show below that there exists $c > 0$ such that $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \frac{c}{|t|} \tag{*} $$ for all $t \in [-1,1]$ such that $f(x,t)$ is nonzero for all $x \in [0,1]$. Therefore, $$ \exp \left( -\int_0^1 \frac{1}{f(x,t)} dx \right) \leq \exp \left( -\frac{c}{|t|} \right) $$ for such $t$. Therefore, $$ G(f) \leq \exp \left( -\frac{c}{|t|} \right) $$ for all nonzero $t \in [-1,1]$, since either $G(f) = 0$, or $G(f)$ is as in the previous equation. It follows that the first derivative of $G(f)$ exists and is zero at $t = 0$. 

Can the argument be adapted to handle the higher derivatives?

Proof of the inequality (*):

Let $x_0 \in [0,1]$ be such that $f(x_0,0) = 0$. Then, by smoothness, there exists $C > 0$ such that $$ f(x,t) \leq C (t^2 + (x - x_0)^2) $$ for every $t \in [-1,1]$ and $x \in [0,1]$. The squares comes from the fact that $f$ is assumed to be a non-negative smooth function, so $\partial_x f(x_0,0) = \partial_t f(x_0,0) = 0$. In particular, if $|x - x_0| \leq |t| \leq 1$, then $$ f(x,t) \leq 2 C t^2 . $$ Choose $t \in [-1,1]$ such that $f(x,t)$ has no zeros for $x \in [0,1]$. Integrating, we get $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \int_{\max(x_0 - |t|,0)}^{\min(x_0 + |t|,1)} \frac{1}{f(x,t)} dx \geq \int_{\max(x_0 - |t|,0)}^{\min(x_0 + |t|,1)} \frac{1}{2 C t^2} dx \geq \frac{|t|}{2 C t^2} = \frac{c}{|t|}, $$ as claimed. The last inequality comes from the fact that the interval of integration has width at least $|t|$.

That $f$ is non-negative is crucial here: the argument given above does not work if $f$ can be real valued, and this can be seen explicitly for the example $f(x,t) = x-t$. However, in such a situation one could simply replace the integrand $1/f$ with $1 / f^2$.

Willie Wong's idea solves the problem. Here's a sketch of the argument.

Edit: there is a mistake (see bold below), so I'm no longer sure if this works.

Define $G(f)$ to be $\exp(-\int_0^1 dx/f(x))$ if $f$ is non-zero on $[0,1]$, and let $G(f) = 0$ otherwise. Let $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0, \infty))$. Without loss of generality, we'll show that all derivatives of $G(f)$ with respect to $t$ at $t=0$ exist. If $f(x,0)$ is nonzero on $[0,1]$, then this is also true for $t$ in a neighbourhood of $0$, and $G(f)$ is smooth in this neighbourhood by differentiating under the integral.

If $f(x,0)$ has a zero in $[0,1]$, then $G(f)$ is zero when $t=0$, and one can show that there exists $c > 0$ such that $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \frac{c}{|t|} \tag{*} $$ for all $t \in [-1,1]$ such that $f(x,t)$ is nonzero for all $x \in [0,1]$. Therefore, $$ \exp \left( -\int_0^1 \frac{1}{f(x,t)} dx \right) \leq \exp \left( -\frac{c}{|t|} \right) $$ for such $t$. Therefore, $$ G(f) \leq \exp \left( -\frac{c}{|t|} \right) $$ for all nonzero $t \in [-1,1]$, since either $G(f) = 0$, or $G(f)$ is as in the previous equation. It follows that the first derivative of $G(f)$ exists and is zero at $t = 0$. Can the argument be adapted to handle the higher derivatives?

The meat is in inequality (*), which is proved using a Taylor approximation to $f$ in the $t$ variable.

Clarification on equation (*):

Let $g(t) = \min_{x\in [0,1]} f(x,t)$ (recall that $f$ is by assumption nonnegative). Since $[0,1]$ is compact and $f$ is continuous, we have that $g$ is continuous. Therefore $g^{-1}(0)$ is a closed set. The argument above the cut shows differentiability of $G(f)(t)$ away from the boundary $\partial g^{-1}(0)$.

Suppose $t = 0$ is at the boundary, such that $g(t) > 0$ in some small interval $(0,t_0)$. Now let $x_0 \in [0,1]$ be such that $f(x_0,0) = 0$. Then by smoothness there exists $\epsilon > 0$ and $C > 0$ such that

$$ f < C (t^2 + (x - x_0)^2)$$

for every $t \in [0,\min(t_0,\epsilon)]$ and $x\in (x - \epsilon, x+\epsilon)\cap [0,1]$. The squares comes from the fact that $f$ is assumed to be a non-negative smooth function, so $\partial_xf (x_0,0) = \partial_t f(x_0,0) = 0$. So integrating we get

$$ \int_0^1 \frac{1}{f} dx \geq \int_{\max(x_0 - t,0)}^{\min(x_0 + t,1)} \frac{1}{f} dx \gtrsim \frac{t}{t^2} = \frac{1}{t} $$

as claimed.

That $f$ is non-negative is crucial here: the argument given above does not work if $f$ can be real valued, and this can be seen explicitly for the example $f(x,t) = x-t$. However, in such a situation we should demand instead of $1/f$ in the integrand, that we use $1 / f^2$.

Willie Wong's idea solves the problem. Here's a sketch of the argument.

Edit: there is a mistake (see bold below), so I'm no longer sure if this works.

Define $G(f)$ to be $\exp(-\int_0^1 dx/f(x))$ if $f$ is non-zero on $[0,1]$, and let $G(f) = 0$ otherwise. Let $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0, \infty))$. Without loss of generality, we'll show that all derivatives of $G(f)$ with respect to $t$ at $t=0$ exist. If $f(x,0)$ is nonzero on $[0,1]$, then this is also true for $t$ in a neighbourhood of $0$, and $G(f)$ is smooth in this neighbourhood by differentiating under the integral.

If $f(x,0)$ has a zero in $[0,1]$, then $G(f)$ is zero when $t=0$, and one can show that there exists $c > 0$ such that $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \frac{c}{|t|} \tag{*} $$ for all $t \in [-1,1]$ such that $f(x,t)$ is nonzero for all $x \in [0,1]$. Therefore, $$ \exp \left( -\int_0^1 \frac{1}{f(x,t)} dx \right) \leq \exp \left( -\frac{c}{|t|} \right) $$ for such $t$. Therefore, $$ G(f) \leq \exp \left( -\frac{c}{|t|} \right) $$ for all nonzero $t \in [-1,1]$, since either $G(f) = 0$, or $G(f)$ is as in the previous equation. It follows that the first derivative of $G(f)$ exists and is zero at $t = 0$. Can the argument be adapted to handle the higher derivatives?

The meat is in inequality (*), which is proved using a Taylor approximation to $f$ in the $t$ variable.

Clarification on equation (*):

Let $g(t) = \min_{x\in [0,1]} f(x,t)$ (recall that $f$ is by assumption nonnegative). Since $[0,1]$ is compact and $f$ is continuous, we have that $g$ is continuous. Therefore $g^{-1}(0)$ is a closed set. The argument above the cut shows differentiability of $G(f)(t)$ away from the boundary $\partial g^{-1}(0)$.

Suppose $t = 0$ is at the boundary, such that $g(t) > 0$ in some small interval $(0,t_0)$. Now let $x_0 \in [0,1]$ be such that $f(x_0,0) = 0$. Then by smoothness there exists $\epsilon > 0$ and $C > 0$ such that

$$ f < C (t^2 + (x - x_0)^2)$$

for every $t \in [0,\min(t_0,\epsilon)]$ and $x\in (x_0 - \epsilon, x_0+\epsilon)\cap [0,1]$. The squares comes from the fact that $f$ is assumed to be a non-negative smooth function, so $\partial_xf (x_0,0) = \partial_t f(x_0,0) = 0$. So integrating we get

$$ \int_0^1 \frac{1}{f} dx \geq \int_{\max(x_0 - t,0)}^{\min(x_0 + t,1)} \frac{1}{f} dx \gtrsim \frac{t}{t^2} = \frac{1}{t} $$

as claimed.

That $f$ is non-negative is crucial here: the argument given above does not work if $f$ can be real valued, and this can be seen explicitly for the example $f(x,t) = x-t$. However, in such a situation we should demand instead of $1/f$ in the integrand, that we use $1 / f^2$.

Willie Wong's idea solves the problem. Here's a sketch of the argument.

Define $G(f)$ to be $\exp(-\int_0^1 dx/f(x))$ if $f$ is non-zero on $[0,1]$, and let $G(f) = 0$ otherwise. Let $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0, \infty))$. Without loss of generality, we'll show that all derivatives of $G(f)$ with respect to $t$ at $t=0$ exist. If $f(x,0)$ is nonzero on $[0,1]$, then this is also true for $t$ in a neighbourhood of $0$, and $G(f)$ is smooth in this neighbourhood by differentiating under the integral.

If $f(x,0)$ has a zero in $[0,1]$, then $G(f)$ is zero when $t=0$, and one can show that there exists $c > 0$ such that $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \frac{c}{|t|} \tag{*} $$ for all $t \in [-1,1]$ such that $f(x,t)$ is nonzero for all $x \in [0,1]$. Therefore, $$ \exp \left( -\int_0^1 \frac{1}{f(x,t)} dx \right) \leq \exp \left( -\frac{c}{|t|} \right) $$ for such $t$. Therefore, $$ G(f) \leq \exp \left( -\frac{c}{|t|} \right) $$ for all nonzero $t \in [-1,1]$, since either $G(f) = 0$, or $G(f)$ is as in the previous equation. It follows that all derivatives of $G(f)$ exist and are zero at $t = 0$.

The meat is in inequality (*), which is proved using a Taylor approximation to $f$ in the $t$ variable.

Clarification on equation (*):

Let $g(t) = \min_{x\in [0,1]} f(x,t)$ (recall that $f$ is by assumption nonnegative). Since $[0,1]$ is compact and $f$ is continuous, we have that $g$ is continuous. Therefore $g^{-1}(0)$ is a closed set. The argument above the cut shows differentiability of $G(f)(t)$ away from the boundary $\partial g^{-1}(0)$.

Suppose $t = 0$ is at the boundary, such that $g(t) > 0$ in some small interval $(0,t_0)$. Now let $x_0 \in [0,1]$ be such that $f(x_0,0) = 0$. Then by smoothness there exists $\epsilon > 0$ and $C > 0$ such that

$$ f < C (t^2 + (x - x_0)^2)$$

for every $t \in [0,\min(t_0,\epsilon)]$ and $x\in (x - \epsilon, x+\epsilon)\cap [0,1]$. The squares comes from the fact that $f$ is assumed to be a non-negative smooth function, so $\partial_xf (x_0,0) = \partial_t f(x_0,0) = 0$. So integrating we get

$$ \int_0^1 \frac{1}{f} dx \geq \int_{\max(x_0 - t,0)}^{\min(x_0 + t,1)} \frac{1}{f} dx \gtrsim \frac{t}{t^2} = \frac{1}{t} $$

as claimed.

That $f$ is non-negative is crucial here: the argument given above does not work if $f$ can be real valued, and this can be seen explicitly for the example $f(x,t) = x-t$. However, in such a situation we should demand instead of $1/f$ in the integrand, that we use $1 / f^2$.

Willie Wong's idea solves the problem. Here's a sketch of the argument.

Edit: there is a mistake (see bold below), so I'm no longer sure if this works.

Define $G(f)$ to be $\exp(-\int_0^1 dx/f(x))$ if $f$ is non-zero on $[0,1]$, and let $G(f) = 0$ otherwise. Let $f(x,t) \in C^\infty(\mathbb{R} \times \mathbb{R}, [0, \infty))$. Without loss of generality, we'll show that all derivatives of $G(f)$ with respect to $t$ at $t=0$ exist. If $f(x,0)$ is nonzero on $[0,1]$, then this is also true for $t$ in a neighbourhood of $0$, and $G(f)$ is smooth in this neighbourhood by differentiating under the integral.

If $f(x,0)$ has a zero in $[0,1]$, then $G(f)$ is zero when $t=0$, and one can show that there exists $c > 0$ such that $$ \int_0^1 \frac{1}{f(x,t)} dx \geq \frac{c}{|t|} \tag{*} $$ for all $t \in [-1,1]$ such that $f(x,t)$ is nonzero for all $x \in [0,1]$. Therefore, $$ \exp \left( -\int_0^1 \frac{1}{f(x,t)} dx \right) \leq \exp \left( -\frac{c}{|t|} \right) $$ for such $t$. Therefore, $$ G(f) \leq \exp \left( -\frac{c}{|t|} \right) $$ for all nonzero $t \in [-1,1]$, since either $G(f) = 0$, or $G(f)$ is as in the previous equation. It follows that the first derivative of $G(f)$ exists and is zero at $t = 0$. Can the argument be adapted to handle the higher derivatives?

The meat is in inequality (*), which is proved using a Taylor approximation to $f$ in the $t$ variable.

Clarification on equation (*):

Let $g(t) = \min_{x\in [0,1]} f(x,t)$ (recall that $f$ is by assumption nonnegative). Since $[0,1]$ is compact and $f$ is continuous, we have that $g$ is continuous. Therefore $g^{-1}(0)$ is a closed set. The argument above the cut shows differentiability of $G(f)(t)$ away from the boundary $\partial g^{-1}(0)$.

Suppose $t = 0$ is at the boundary, such that $g(t) > 0$ in some small interval $(0,t_0)$. Now let $x_0 \in [0,1]$ be such that $f(x_0,0) = 0$. Then by smoothness there exists $\epsilon > 0$ and $C > 0$ such that

$$ f < C (t^2 + (x - x_0)^2)$$

for every $t \in [0,\min(t_0,\epsilon)]$ and $x\in (x - \epsilon, x+\epsilon)\cap [0,1]$. The squares comes from the fact that $f$ is assumed to be a non-negative smooth function, so $\partial_xf (x_0,0) = \partial_t f(x_0,0) = 0$. So integrating we get

$$ \int_0^1 \frac{1}{f} dx \geq \int_{\max(x_0 - t,0)}^{\min(x_0 + t,1)} \frac{1}{f} dx \gtrsim \frac{t}{t^2} = \frac{1}{t} $$

as claimed.

That $f$ is non-negative is crucial here: the argument given above does not work if $f$ can be real valued, and this can be seen explicitly for the example $f(x,t) = x-t$. However, in such a situation we should demand instead of $1/f$ in the integrand, that we use $1 / f^2$.