An example with negative $dx$
When using linear approximations, $x$ doesn't have to be bigger
than $a$.
Here is an example where $x$ is slightly less than $a$. The
work is the same as when $x$ is larger than $a$:
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 ${f'(x) = \frac{d}{dx} \left(\ln(x)\right)= \frac{1}{x}}$,
we have $f'(a) = {\frac{1}{10}}$.
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*}
