# 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!

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If $0 \lt a \le b$ then by symmetry $\int_{-b}^{-a} \frac{1}{x} \mathrm{d}x = -\int_{a}^{b} \frac{1}{x} \mathrm{d}x = \log_e(a)-\log_e(b)$ –  Henry Dec 30 '11 at 22:42

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 $\int_a^0\frac{1}{x}\,dx$.

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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!