Solution
= Solution
Factor $1+e^x=2e^{x/2}\cosh(x/2)$. This gives
$$
\ln(1+e^x)=\ln2+\frac x2+\ln\cosh\frac x2.
$$
The last <function> is even. Its <Taylor series> begins $\ln\cosh(x/2)=x^2/8+O(x^4)$, since $\cosh(x/2)=1+x^2/8+O(x^4)$ and $\ln(1+w)=w+O(w^2)$. Therefore
$$
\boxed{\ln(1+e^x)=\ln2+\frac x2+\frac{x^2}{8}+0x^3+O(x^4)}.
$$
In particular, the cubic coefficient is zero; the identity explains its cancellation before any lengthy differentiation.