From Braddick's The Physics of Experimental Method (1963):"It should be remembered that the number of significant figures in a number $y$, and one derived from it, say $e^y$, are not always the same. Thus if $y=1.32\times 10^{-2}$ (three significant figures) then $e^y=1.1034$ (five significant figures)".
What is the rationale behind the number of significant figures in the above example ?
Following does not address the query: Number of significant figures
If $y=1.32\times 10^{-2}$, then this means that exact value of $y$ can be anywhere in the interval $[1.315\times 10^{-2}, 1.325\times 10^{-2}].$ If you now exponentiate the endpoints, then, since the exponential function is monotonic, you'll get the interval where $\exp(y)$ can be: $[1.01323684148769,1.01333817023819].$ Now what's the difference between these endpoints? It's about $0.0001$. This is the absolute error of $\exp(y)$. You can now easily see that we know this exponent to (a bit less than) 5 significant figures.
Remember that one representation for the exponential function is
$$ e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots $$
Suppose I know $y$ to only one significant figure, $y = 2\times10^{-12}$. I immediately know $e^y$ to twelve significant figures:
\begin{align} \exp({2\times10^{-12}}) &= 1 + 2\times10^{-12} + \frac{2^2}{2!}\times10^{-24} + \cdots \\& \approx 1.000\,000\,000\,002 \end{align}
