Why can natural logarithm receive negative $x$? (in the context of derivatives)

Question

In our class we had this formula for $x<0$: $$ \ln(-x)' = \frac{1}{-x} \cdot (-x)' = \frac{1}{x} $$ The conclusion is that: $$ \int\frac{1}{x}dx = \ln|x| + c, x \neq 0 $$

I don't understand how can $\ln$ receive negative numbers though?

Answer 1

It doesn't. The first formula is for negative $x$, this means that $-x$ is a positive number.

The second formula (the conclusion) is because the first we have that:

$$D\ln |x| = \begin{cases} D\ln x = 1/x & \text{ if } x > 0\\ D\ln (-x) = 1/x & \text { if } x < 0\\ \end{cases}$$

where the second case is from the formula above, the first case is supposedly already known.