已知 $f(x) = \ln\left(1+e^{-x}\right)$。

1. Expand $e^{-x}$ using the standard series up to the $x^2$ term.

$1 + e^{-x} = 1 + \left(1 - x + \frac{(-x)^2}{2!} + \dots\right) \approx 2 - x + \frac{x^2}{2}$

2. Substitute the expansion into the given function and factorise the argument to prepare for the logarithmic standard series.

$f(x) = \ln(1+e^{-x}) \approx \ln\left(2 - x + \frac{x^2}{2}\right) = \ln\left[2\left(1 + \left(-\frac{x}{2} + \frac{x^2}{4}\right)\right)\right]$

3. Apply the product law of logarithms to split the expression.

$\ln\left[2\left(1 + \left(-\frac{x}{2} + \frac{x^2}{4}\right)\right)\right] = \ln 2 + \ln\left(1 + \left(-\frac{x}{2} + \frac{x^2}{4}\right)\right)$

4. Expand $\ln(1+X)$ using the standard series up to the $x^2$ term, simplify, and identify $a$, $b$, and $c$.

Using $\ln(1+X) \approx X - \frac{X^2}{2}$:
$f(x) \approx \ln 2 + \left(-\frac{x}{2} + \frac{x^2}{4}\right) - \frac{1}{2}\left(-\frac{x}{2} + \frac{x^2}{4}\right)^2$
$= \ln 2 - \frac{x}{2} + \frac{x^2}{4} - \frac{1}{2}\left(\frac{x^2}{4}\right) + \dots$
$= \ln 2 - \frac{1}{2}x + \frac{1}{8}x^2$
Comparing coefficients with $\ln a + bx + cx^2$, we obtain $a = 2$, $b = -\frac{1}{2}$, and $c = \frac{1}{8}$.

1. Derive the constants $a$, $b$, and $c$ for the Maclaurin series of $\mathrm{f}(x)$ as required from part (i).

From part (i), we find the derivatives at $x=0$:
$\mathrm{f}(x) = \ln(1+e^{-x}) \implies \mathrm{f}(0) = \ln 2$
$\mathrm{f}'(x) = \dfrac{-e^{-x}}{1+e^{-x}} = \dfrac{-1}{e^x+1} \implies \mathrm{f}'(0) = -\dfrac{1}{2}$
$\mathrm{f}''(x) = \dfrac{d}{dx}\left(-(e^x+1)^{-1}\right) = (e^x+1)^{-2}e^x \implies \mathrm{f}''(0) = \dfrac{1}{4}$
The Maclaurin series is $\mathrm{f}(0) + \mathrm{f}'(0)x + \dfrac{\mathrm{f}''(0)}{2!}x^2$:
$\mathrm{f}(x) \approx \ln 2 - \dfrac{1}{2}x + \dfrac{1}{8}x^2$

2. Express the target expression in terms of $\mathrm{f}'(x)$.

Observe the first derivative of $\mathrm{f}(x)$:
$\mathrm{f}'(x) = \dfrac{-e^{-x}}{1+e^{-x}}$
Multiplying the numerator and denominator by $e^x$:
$\mathrm{f}'(x) = \dfrac{-1}{e^x+1}$
Rearranging this gives a direct relationship with the target expression:
$\dfrac{1}{1+e^x} = -\mathrm{f}'(x)$

3. Differentiate the Maclaurin series of $\mathrm{f}(x)$ to find the required expansion.

Differentiating the polynomial approximation of $\mathrm{f}(x)$ with respect to $x$:
$\mathrm{f}'(x) \approx \dfrac{d}{dx}\left(\ln 2 - \dfrac{1}{2}x + \dfrac{1}{8}x^2\right)$
$\mathrm{f}'(x) \approx -\dfrac{1}{2} + \dfrac{1}{4}x$
Therefore, substituting into $\dfrac{1}{1+e^x} = -\mathrm{f}'(x)$:
$\dfrac{1}{1+e^x} \approx -\left(-\dfrac{1}{2} + \dfrac{1}{4}x\right) = \dfrac{1}{2} - \dfrac{1}{4}x$