Question 14

We need to find $\frac{d}{dx} [f(g^{-1}(x))]$ at $x=2$. By chain rule, this is $f'(g^{-1}(2)) \cdot (g^{-1})'(2)$.

1. Find $g^{-1}(2)$: Let $g(x) = 2 \implies \frac{4}{1 + e^{-2x}} = 2 \implies 1 + e^{-2x} = 2 \implies e^{-2x} = 1 \implies x = 0$. So $g^{-1}(2) = 0$.
2. Find $(g^{-1})'(2)$: Using the inverse function derivative rule, $(g^{-1})'(2) = \frac{1}{g'(g^{-1}(2))} = \frac{1}{g'(0)}$. $g'(x) = \frac{d}{dx} [4(1 + e^{-2x})^{-1}] = -4(1 + e^{-2x})^{-2} \cdot (-2e^{-2x}) = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$. $g'(0) = \frac{8(1)}{(1+1)^2} = \frac{8}{4} = 2$. So $(g^{-1})'(2) = \frac{1}{2} = 0.5$.
3. Find $f'(g^{-1}(2)) = f'(0)$: $f'(x) = \frac{d}{dx} [\log_e(x^2 + 2x + 4)] = \frac{2x+2}{x^2+2x+4}$. $f'(0) = \frac{2}{4} = 0.5$.
4. Final result: $0.5 \cdot 0.5 = 0.25$.