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 $x-a = -0.05$.
Since ${f'(x) = \frac{d}{dx} \left(\ln(x)\right)= \frac{1}{x}}$, we have $f'(a) = {\frac{1}{10}}$.
Our linearization is then: $$L(x) = 2.30258 + \frac{1}{10}(x-10).$$ 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*}