(ii) $\frac{\mathrm{e}^x}{1+\mathrm{e}^x}=\frac{1}{2}+\frac{1}{4} x+\ldots$

(i) $y=\ln \left(1+\mathrm{e}^x\right)$
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{e}^x}{1+\mathrm{e}^x}$
$\left(1+\mathrm{e}^x\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{e}^x$
$\left(1+\mathrm{e}^x\right) \frac{\mathrm{d} y}{\mathrm{~s}^x}-\mathrm{e}^x=0$ (shown)

(ii) $\left(1+\mathrm{e}^x\right) \frac{\mathrm{d}^2 y}{d x^2}+\mathrm{e}^x \frac{\mathrm{d} y}{\mathrm{~d} x}-\mathrm{e}^x=0$
$\left(1+\mathrm{e}^x\right) \frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}+\mathrm{e}^x \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\mathrm{e}^x \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\mathrm{e}^x \frac{\mathrm{d} y}{\mathrm{~d} x}-\mathrm{e}^x=0$
$\left(1+\mathrm{e}^x\right) \frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}+2 \mathrm{e}^x \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\mathrm{e}^x \frac{\mathrm{d} y}{\mathrm{~d} x}-\mathrm{e}^x=0$
When $x=0, y=\ln 2$
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{e}^0}{1+\mathrm{e}^0}=\frac{1}{2}$
$\left(1+\mathrm{e}^0\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\mathrm{e}^0\left(\frac{1}{2}\right)-\mathrm{e}^0=0$
$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{1}{4}$

By Maclaurin’s series,
$$\begin{array}{l}
y=\ln 2+\frac{1}{2} x+\frac{\frac{1}{4}}{2 !} x^2+\cdots \\
y=\ln 2+\frac{1}{2} x+\frac{1}{8} x^2+\cdots \\
\text { Since } y=\ln \left(1+\mathrm{e}^x\right), \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{e}^x}{1+\mathrm{e}^x} \\
\frac{\mathrm{e}^x}{1+\mathrm{e}^x}=\frac{1}{2}+\frac{1}{4} x+\ldots
\end{array}$$

(iii) Using MF26,
$$
\begin{array}{l}
\frac{\mathrm{e}^x}{1+\mathrm{e}^x}=\mathrm{e}^x\left(1+\mathrm{e}^x\right)^{-1} \\
=(1+x+\cdots)(1+1+x+\cdots)^{-1} \\
\approx(1+x)\left(2^{-1}\left(1+\frac{x}{2}\right)^{-1}\right) \\
=\frac{1}{2}(1+x)\left(1+\frac{x}{2}+\cdots\right) \\
=\frac{1}{2}\left(1+x+\frac{x}{2}+\cdots\right) \\
=\frac{1}{2}+\frac{1}{4} x+\ldots
\end{array}
$$