Why is the derivative of $\ln{x}$ defined for negative values?

Question

The derivative of $\ln{x}$ is $\frac{1}{x}$. However, we know that $\ln{x}$ is defined only for $x \gt 0$, yet we know that $\frac{1}{x}$ is defined for $x \lt 0$ as well. How can this be? I was under the impression that a derivative can only be defined when the function is defined, since the function is part of the definition of the derivative.

Answer 1

The derivative of $\ln x$ is $\frac 1x$, if $x$ is positive. If $x$ is negative, the neither $\ln x$ nor its derivative are defined. The fact that the expression $\frac 1x$ is also valid for negative values of $x$ does not mean that the derivative is valid for negative values of $x$.

In other words, it's just a lucky coincidence of how we chose to write expressions. By sheer luck, it happens that the expression we use to describe the derivative of $\ln x$ is valid for negative values of $x$.

We could easily choose a different set of expressions. For example, from now on, I decide that $\mathfrak D$ is defined as follows:

For any positive real number $a$ and any positive real number $b$, we define $a\mathfrak D b$ to be the unique solution to the equation $bx=a$. If either $a$ and $b$ is non-positive, then $a\mathfrak Db$ is not defined.

Once we define this symbol, we can prove that the derivative of $\ln x$ is $1\mathfrak Dx$, and that expression is only defined for positive values of $x$, so the issue you described before is no longer there.

Answer 2

A function can only have a derivative in points where it is defined, so $\log x$, taken as a real function, does not have a derivative for $x<0$. The fact that the expression that defines the derivative for $x>0$ is also defined for $x<0$ does not make it the derivate in that case.