Why isn't $\frac{\mathrm{d} }{\mathrm{d} x} \ln(x)$ specified as $\frac{1}{x},x>0$? As I understand, $\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x}\ln(x)\end{eqnarray} $ is generally specified as $\begin{eqnarray} \frac{1}{x} \end{eqnarray}$.  Would it be more appropriate to state it as $\begin{eqnarray} \frac{1}{x}, x>0\end{eqnarray}$ since $\ln(x)$ is undefined for $x\leq0$?  If not, why not?  In addition, what does this imply about the indefinite integral of or the definite integral with negative limits of integration of $\begin{eqnarray} \frac{1}{x} \end{eqnarray}$?
Thank you!
 A: Another answer.  Some strange people use complex numbers, not just real numbers.  For them, log(x) makes sense other than for $x>0$.  And for them its derivative is $1/x$.  More importantly: for them, $\int (1/x)\,dx = \log(|x|)+C$ is WRONG!
A: Whenever we write something like $\ln(x)$, we are implicitly asserting that $x$ is restricted to those $x$s for which the expression makes sense. 
When we write $$\frac{d}{dx}\ln(x) = \frac{1}{x}$$ we are then implicitly and automatically saying that $x$ is positive for the equation to make sense. As for integrals, that's why $$\int\frac{1}{x}\,dx = \ln|x|+C,$$
with the absolute values: note that
$$\frac{d}{dx}\ln(-x) = \frac{1}{-x}(-x)' = \frac{1}{x}$$
as well by the Chain Rule, so that
$$\frac{d}{dx}\ln|x| = \frac{1}{x}.$$
For a definite integral to make sense, you usually need the function defined on the entire interval of integration (at least, for the usual Riemann integrals), so $\int_a^b\frac{1}{x}\,dx$ cannot have $a\lt 0 \lt b$ and make sense. Either $0\lt a\lt b$ or $a\lt b\lt 0$.
You can try to do an integral $\int_a^b \frac{1}{x}\,dx$ with $a\lt 0\lt b$ as an improper integral, but you will find that the integral does not converge; neither do $\int_0^b\frac{1}{x}\,dx$ nor $\int_a^0\frac{1}{x}\,dx$. 
