Step-by-step Derivations

1. Understand the Goal We need to find the value of the derivative of the composite function $f \circ g^{-1}$ at $x=2$. This is denoted as $(f \circ g^{-1})'(2)$.

2. Apply the Chain Rule Using the chain rule for differentiation, the derivative of the composite function $h(x) = f(g^{-1}(x))$ is given by: $h'(x) = (f \circ g^{-1})'(x) = f'(g^{-1}(x)) \cdot (g^{-1})'(x)$ To find the value at $x=2$, we need to evaluate: $(f \circ g^{-1})'(2) = f'(g^{-1}(2)) \cdot (g^{-1})'(2)$ This requires us to find three values: $g^{-1}(2)$, $f'(g^{-1}(2))$, and $(g^{-1})'(2)$.

3. Calculate $g^{-1}(2)$ Let $y = g^{-1}(2)$. By the definition of an inverse function, this means $g(y) = 2$. We are given the function $g(x)=\frac{4}{1+e^{-2 x}}$. $g(y) = \frac{4}{1+e^{-2 y}} = 2$ Solving for $y$: $4 = 2(1+e^{-2 y})$ $2 = 1+e^{-2 y}$ $1 = e^{-2 y}$ Taking the natural logarithm of both sides: $\ln(1) = -2y$ $0 = -2y$ $y = 0$ So, we have $g^{-1}(2) = 0$.

4. Calculate $f'(g^{-1}(2))$ First, we find the derivative of $f(x) = \log_e(x^2+2x+4)$. Using the chain rule, $f'(x) = \frac{1}{x^2+2x+4} \cdot \frac{d}{dx}(x^2+2x+4)$. $f'(x) = \frac{2x+2}{x^2+2x+4}$ Now, we evaluate this at $x = g^{-1}(2) = 0$. $f'(g^{-1}(2)) = f'(0) = \frac{2(0)+2}{0^2+2(0)+4} = \frac{2}{4} = \frac{1}{2}$

5. Calculate $(g^{-1})'(2)$ We can use the formula for the derivative of an inverse function: $(g^{-1})'(a) = \frac{1}{g'(g^{-1}(a))}$. For $a=2$, we have $(g^{-1})'(2) = \frac{1}{g'(g^{-1}(2))}$. From Step 3, we know $g^{-1}(2) = 0$. So, we need to find $g'(0)$.

First, find the derivative of $g(x) = \frac{4}{1+e^{-2 x}} = 4(1+e^{-2x})^{-1}$. Using the chain rule: $g'(x) = 4 \cdot (-1)(1+e^{-2x})^{-2} \cdot \frac{d}{dx}(1+e^{-2x})$ $g'(x) = -4(1+e^{-2x})^{-2} \cdot (e^{-2x} \cdot -2)$ $g'(x) = \frac{8e^{-2x}}{(1+e^{-2x})^2}$ Now, evaluate $g'(x)$ at $x=0$: $g'(0) = \frac{8e^{-2(0)}}{(1+e^{-2(0)})^2} = \frac{8e^0}{(1+e^0)^2} = \frac{8(1)}{(1+1)^2} = \frac{8}{2^2} = \frac{8}{4} = 2$ Using the inverse function derivative formula: $(g^{-1})'(2) = \frac{1}{g'(0)} = \frac{1}{2}$

6. Final Calculation Now we can substitute the values we found back into the chain rule expression from Step 2: $(f \circ g^{-1})'(2) = f'(g^{-1}(2)) \cdot (g^{-1})'(2)$ $(f \circ g^{-1})'(2) = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{4}$ As a decimal, this is $0.25$.

7. Conclusion The value of the derivative of the composite function $f \circ g^{-1}$ at $x=2$ is $0.25$. This value lies in the range $0.2$ to $0.3$.