Function $ f $:

$ f(x) = \log_e\left(x^2 + 2x + 4\right) = \log_e\left[(x+1)^2 + 3\right] $

Function $ g $:

$ g(x) = \frac{4}{1 + e^{-2x}} $

To find the inverse of $ g $, $ g^{-1} $, we solve for $ x $:

Start with $ y = \frac{4}{1 + e^{-2x}} $.

Rearranging gives $ 1 + e^{-2x} = \frac{4}{y} $.

Solving for $ e^{-2x} $, we have $ e^{-2x} = \frac{4}{y} - 1 $.

Taking the natural logarithm, we find $ -2x = \ln\left(\frac{4}{y} - 1\right) $.

Thus, we solve for $ x $:

$ x = -\frac{1}{2} \ln\left(\frac{4-y}{y}\right) = \frac{1}{2} \ln\left(\frac{y}{4-y}\right) $

So, the inverse function:

$ g^{-1}(x) = \frac{1}{2} \ln\left(\frac{x}{4-x}\right) $

Now, let's evaluate $ g^{-1}(2) $:

$ g^{-1}(2) = \frac{1}{2} \ln\left(\frac{2}{4-2}\right) = \frac{1}{2} \ln(1) = 0 $

Thus, $ f(g^{-1}(x)) = f\left(\frac{1}{2} \ln\left(\frac{x}{4-x}\right)\right) $.

Next, we differentiate $ f \circ g^{-1} $:

$ \frac{d}{dx}\left(f(g^{-1}(x))\right) = \frac{d}{dx}\left[\log_e\left(\left(g^{-1}(x) + 1\right)^2 + 3\right)\right] $

Using the chain rule, the derivative is:

$ \frac{1}{\left[\left(g^{-1}(x) + 1\right)^2 + 3\right]} \cdot 2 \left(g^{-1}(x) + 1\right) \cdot \frac{d}{dx}\left(g^{-1}(x)\right) $

Calculate the derivative of $ g^{-1}(x) $:

$ \frac{d}{dx}\left(\frac{1}{2} \ln\left(\frac{x}{4-x}\right)\right) = \frac{1}{2} \cdot \frac{1}{\frac{x}{4-x}} \cdot \left(\frac{(4-x)x' + x(4-x)'}{(4-x)^2}\right) $

Evaluate at $ x = 2 $:

$ g^{-1}(2) = 0 $ means $ \left(g^{-1}(2) + 1\right)^2 + 3 = 4 $.

At $ x = 2 $:

$ \frac{d}{dx} f(g^{-1}(x))\big|_{x=2} = \frac{2}{4} \cdot \frac{2}{4} = \frac{1}{4} = 0.25 $

Therefore, the value of the derivative of the composite function $ f \circ g^{-1} $ at $ x = 2 $ is $ 0.25 $.