Derivative of Composite Function Involving Inverse Function at Specific Point

To find the derivative of the composite function $(f \circ g^{-1})(x) = f(g^{-1}(x))$ at $x = 2$, we use the chain rule:

$\left(f \circ g^{-1}\right)'(x) = f'\left(g^{-1}(x)\right) \cdot \left(g^{-1}\right)'(x)$

Using the formula for the derivative of an inverse function, $\left(g^{-1}\right)'(x) = \frac{1}{g'\left(g^{-1}(x)\right)}$, we get:

$\left(f \circ g^{-1}\right)'(2) = f'\left(g^{-1}(2)\right) \cdot \frac{1}{g'\left(g^{-1}(2)\right)}$

Step 1: Find $g^{-1}(2)$

Let $y = g^{-1}(2)$, which means $g(y) = 2$. Given $g(x) = \frac{4}{1 + e^{-2x}}$:

$\frac{4}{1 + e^{-2y}} = 2 \implies 1 + e^{-2y} = 2 \implies e^{-2y} = 1 \implies y = 0$

Thus, $g^{-1}(2) = 0$.

Step 2: Calculate $f'(0)$

Given $f(x) = \log_e(x^2 + 2x + 4)$, its derivative with respect to $x$ is:

$f'(x) = \frac{2x + 2}{x^2 + 2x + 4}$

Evaluating $f'(x)$ at $x = 0$:

$f'(0) = \frac{2(0) + 2}{0^2 + 2(0) + 4} = \frac{2}{4} = \frac{1}{2}$

Step 3: Calculate $g'(0)$

Given $g(x) = \frac{4}{1 + e^{-2x}}$, its derivative with respect to $x$ is:

$g'(x) = 4 \cdot (-1)(1 + e^{-2x})^{-2} \cdot (-2e^{-2x}) = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$

Evaluating $g'(x)$ at $x = 0$:

$g'(0) = \frac{8e^0}{(1 + e^0)^2} = \frac{8}{(1 + 1)^2} = \frac{8}{4} = 2$

Step 4: Compute the final value

Substituting $g^{-1}(2) = 0$, $f'(0) = \frac{1}{2}$, and $g'(0) = 2$ into our chain rule expression:

$\left(f \circ g^{-1}\right)'(2) = \frac{f'(0)}{g'(0)} = \frac{1/2}{2} = \frac{1}{4} = 0.25$

The value of the derivative at $x = 2$ is $0.25$.