Solution
= Solution
The <second derivative chain rule> gives
$$
f''(x)=\frac{t''(x)}{1+t(x)}-\frac{t'(x)^2}{(1+t(x))^2}.
$$
With $t_0=0$, the constant term vanishes, $f'(0)=t_1\ne0$, and $f''(0)=t_2-t_1^2\ne0$ by the specified case. Thus the first two nonzero terms in the <Taylor expansion> are
$$
\boxed{f(x)=t_1x+\frac{t_2-t_1^2}{2}x^2+O(x^3).}
$$