Example: If $\ln(10) = 2.30258$ (to 5 decimal places), what is
$\ln(9.95)$?
Solution:
In this problem we're working with $f(x) = \ln(x)$, $a=10$, and
$x = 9.95$, so $xa = 0.05$.
Since $\displaystyle{f'(x) = \frac{d}{dx} \big(\ln(x)\big)= \frac{1}{x}}$, we have
$f'(a) = \displaystyle{\frac{1}{10}}$. Our linearization is then:
$$L(x) = 2.30258 + \frac{1}{10}(x10).$$
Plugging in $x=9.95$ gives
$$ \ln(9.95) \approx L(9.95) = 2.30258 + \frac{1}{10}(0.05) = 2.29758$$
In fact, $\ln(9.95) = 2.29757$ to 5 decimal places, so we were off by
0.00001. The linear approximation isn't exact, but it is very, very good.
Here is the same calculation in terms of differentials:
\begin{eqnarray*}
dx &=& 9.95  10 =  0.05 \cr
df &=& \frac{dx}{10} = 0.005 \cr
f(x) &\approx& f(a) + df = 2.30258  0.005 = 2.29758
\end{eqnarray*}
