Let $\mathbb{R}$ denote the set of all real numbers. Let $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to (0,4)$ be functions defined by $f(x) = \log_e(x^{2} + 2x + 4)$, and $g(x) = \frac{4}{1 + e^{-2x}}$. Define the composite function $f \circ g^{-1}$ by $(f \circ g^{-1})(x) = f(g^{-1}(x))$, where $g^{-1}$ is the inverse of the function $g$. Then the value of the derivative of the composite function $f \circ g^{-1}$ at $x = 2$ is ________.

$f(x) = \log_e(x^2+2x+4), \qquad g(x) = \dfrac{4}{1+e^{-2x}}$ $(f \circ g^{-1})(x) = f\big(g^{-1}(x)\big)$ $\text{To find: } \dfrac{d}{dx}(f \circ g^{-1})\Big|_{x=2}$ $1.\ \text{find } g^{-1}(2)$ $2.\ f'(x), \ g'(x)$ $3.\ (f \circ g^{-1})'(2) = \dfrac{f'(g^{-1}(2))}{g'(g^{-1}(2))}$ $(f \circ g^{-1})'(x) = f'(g^{-1}(x))\cdot(g^{-1})'(x)$ $(g^{-1})'(x) = \dfrac{1}{g'(g^{-1}(x))} \ \Rightarrow \ (f \circ g^{-1})'(x) = \dfrac{f'(g^{-1}(x))}{g'(g^{-1}(x))}$ $g(u)=2 \Rightarrow \dfrac{4}{1+e^{-2u}} = 2 \Rightarrow e^{-2u} = 1 \Rightarrow u = 0$ $g^{-1}(2) = 0$ $f'(x) = \dfrac{2x+2}{x^2+2x+4}$ $g'(x) = \dfrac{8e^{-2x}}{(1+e^{-2x})^2}$ $f'(0) = \dfrac{2(0)+2}{0^2+2(0)+4} = \dfrac{2}{4} = \dfrac{1}{2}$ $g'(0) = \dfrac{8e^{0}}{(1+e^{0})^2} = \dfrac{8}{4} = 2$ $(f \circ g^{-1})'(2) = \dfrac{f'(g^{-1}(2))}{g'(g^{-1}(2))} = \dfrac{f'(0)}{g'(0)} = \dfrac{1/2}{2} = \dfrac{1}{4}$ $\boxed{(f \circ g^{-1})'(2) = \dfrac{1}{4}}$