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

$$f(x) = \log_e(x^2 + 2x + 4)$$, $$\quad$$ and $$\quad g(x) = \frac{4}{1 + e^{-2x}}$$.

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 ______.

We have two functions: $$f(x)=\log_e(x^2+2x+4)$$ for all $$x\in\mathbb{R}$$ and $$g(x)=\dfrac{4}{1+e^{-2x}}$$ whose range is $$(0,\,4)$$.

The composite function is$$(f\circ g^{-1})(y)=f\!\bigl(g^{-1}(y)\bigr),\qquad 0\lt y\lt 4.$$We must find its derivative at $$y=2$$.

Step 1: Find the inverse $$g^{-1}$$.
Let $$y=g(x)=\dfrac{4}{1+e^{-2x}}.$$ Multiply: $$y\,(1+e^{-2x})=4\;\;\Longrightarrow\;\;y\,e^{-2x}=4-y.$$ Therefore $$e^{-2x}=\dfrac{4-y}{y}.$$ Taking natural logarithm,
$$-2x=\ln\!\Bigl(\dfrac{4-y}{y}\Bigr)\;\;\Longrightarrow\;\;x=\dfrac12\ln\!\Bigl(\dfrac{y}{4-y}\Bigr).$$ Thus$$g^{-1}(y)=\dfrac12\ln\!\Bigl(\dfrac{y}{4-y}\Bigr),\qquad 0\lt y\lt4.$$ For convenience, write$$x=g^{-1}(y)=\dfrac12\ln\!\Bigl(\dfrac{y}{4-y}\Bigr).$$

Step 2: Differentiate $$f(x)$$.
$$f(x)=\ln(x^2+2x+4)$$
Using $$\dfrac{d}{dx}\ln u=\dfrac{u'}{u}$$,
$$f'(x)=\dfrac{2x+2}{x^2+2x+4}= \dfrac{2(x+1)}{x^2+2x+4}.$$ At $$x=0$$ (value found later),
$$f'(0)=\dfrac{2(0+1)}{0^2+0+4}= \dfrac{2}{4}= \dfrac12.$$

Step 3: Differentiate $$x=g^{-1}(y)$$ with respect to $$y$$.
$$x=\dfrac12\ln\!\Bigl(\dfrac{y}{4-y}\Bigr).$$ Let $$u(y)=\dfrac{y}{4-y}.$$ Then $$x=\dfrac12\ln u.$$ First evaluate $$u'(y)$$ using the quotient rule:
$$u'(y)=\dfrac{(4-y)\cdot1-y(-1)}{(4-y)^2}= \dfrac{4}{(4-y)^2}.$$ Now$$\dfrac{d}{dy}\ln u=\dfrac{u'(y)}{u(y)}=\dfrac{4}{(4-y)^2}\,\bigg/\,\dfrac{y}{4-y}= \dfrac{4}{y(4-y)}.$$ Hence$$\dfrac{dx}{dy}= \dfrac12\cdot\dfrac{4}{y(4-y)}= \dfrac{2}{y(4-y)}.$$ At $$y=2$$,
$$\dfrac{dx}{dy}\bigg|_{y=2}= \dfrac{2}{2\,(4-2)}= \dfrac{2}{4}= \dfrac12.$$

Step 4: Value of $$x$$ when $$y=2$$.
$$x=g^{-1}(2)=\dfrac12\ln\!\Bigl(\dfrac{2}{4-2}\Bigr)=\dfrac12\ln 1=0.$$

Step 5: Apply the Chain Rule.
For the composite $$h(y)=(f\circ g^{-1})(y)=f(x)\,,$$
$$h'(y)=f'(x)\,\dfrac{dx}{dy}.$$ At $$y=2$$ (so $$x=0$$):
$$h'(2)=f'(0)\times\dfrac{dx}{dy}\bigg|_{y=2}= \dfrac12 \times \dfrac12= \dfrac14=0.25.$$

Therefore the derivative of the composite function at $$x=2$$ is $$0.25$$, which lies in the required range $$0.2-0.3$$.