= Solution
Let $f(x)=\cosh\sqrt x$ near $x=2$. The <chain rule> gives
$$
f'(x)=\frac{\sinh\sqrt x}{2\sqrt x},\qquad
f''(x)=\frac{\cosh\sqrt x}{4x}-\frac{\sinh\sqrt x}{4x^{3/2}}.
$$
Substitute $x=2$ into the first three terms of the <Taylor expansion>:
$$
\boxed{\cosh\sqrt x
=\cosh\sqrt2+\frac{\sinh\sqrt2}{2\sqrt2}(x-2)
+\left(\frac{\cosh\sqrt2}{16}-\frac{\sinh\sqrt2}{16\sqrt2}\right)(x-2)^2
+O((x-2)^3).}
$$
The quadratic coefficient is $f''(2)/2$, not $f''(2)$. The apparent square root creates no local difficulty at the positive expansion point.
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