Question

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

Hint:

To find the derivative of a composite function involving an inverse, use the formula \(\frac{d}{dx} f(g^{-1}(x)) = \frac{f'(g^{-1}(x))}{g'(g^{-1}(x))}\). Always find the inverse value first, then differentiate each function separately.

Answer 1

Step 1: Recall that if \(h = f \circ g^{-1}\), then by the chain rule, \[ h'(x) = \frac{d}{dx} f(g^{-1}(x)) = f'(g^{-1}(x)) \cdot \frac{d}{dx} g^{-1}(x). \] Using the formula for derivative of inverse function, \[ \frac{d}{dx} g^{-1}(x) = \frac{1}{g'(g^{-1}(x))}. \] Thus, \[ h'(x) = \frac{f'(g^{-1}(x))}{g'(g^{-1}(x))}. \]
Step 2: We need to find \(h'(2)\). First find \(g^{-1}(2)\). Given, \[ g(x) = \frac{4}{1 + e^{-2x}}. \] Set \(y = g(x) = 2\), \[ 2 = \frac{4}{1 + e^{-2x}} \implies 1 + e^{-2x} = 2, \] \[ e^{-2x} = 1, \] \[ -2x = 0 \implies x = 0. \] So, \[ g^{-1}(2) = 0. \]
Step 3: Calculate \(f'(x)\). \[ f(x) = \log_e (x^2 + 2x + 4). \] Using chain rule, \[ f'(x) = \frac{1}{x^2 + 2x + 4} \cdot \frac{d}{dx} (x^2 + 2x + 4) = \frac{2x + 2}{x^2 + 2x + 4}. \] Evaluate at \(x = g^{-1}(2) = 0\): \[ f'(0) = \frac{2(0) + 2}{0^2 + 2(0) + 4} = \frac{2}{4} = \frac{1}{2}. \]
Step 4: Calculate \(g'(x)\). \[ g(x) = \frac{4}{1 + e^{-2x}} = 4 \left(1 + e^{-2x}\right)^{-1}. \] Differentiating, \[ g'(x) = 4 \cdot (-1) \cdot \left(1 + e^{-2x}\right)^{-2} \cdot \frac{d}{dx} (1 + e^{-2x}). \] \[ \frac{d}{dx} (1 + e^{-2x}) = -2 e^{-2x}. \] Thus, \[ g'(x) = 4 \cdot (-1) \cdot \left(1 + e^{-2x}\right)^{-2} \cdot (-2 e^{-2x}) = 8 \cdot \frac{e^{-2x}}{(1 + e^{-2x})^2}. \] Evaluate at \(x = 0\): \[ g'(0) = 8 \cdot \frac{e^{0}}{(1 + e^{0})^2} = 8 \cdot \frac{1}{(1 + 1)^2} = 8 \cdot \frac{1}{4} = 2. \] Step 5: Finally, compute \[ h'(2) = \frac{f'(g^{-1}(2))}{g'(g^{-1}(2))} = \frac{f'(0)}{g'(0)} = \frac{\frac{1}{2}}{2} = \frac{1}{4}. \] Since this value \(\frac{1}{4}\) is not listed among the options, there might be a typo in the question or options. But mathematically, the value of the derivative is \(\frac{1}{4}\).