# Is ln(x) differentiable at negative x-axis

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1. Jan 12, 2016

### Miraj Kayastha

Since lnx is defined for positive x only shouldnt the derivative of lnx be 1/x, where x is positive. My books does not specify that x must be positive, so is lnx differentiable for all x?

2. Jan 12, 2016

### Staff: Mentor

No. If a function isn't defined on some interval, its derivative isn't defined there, either. The real function ln(x) is defined only for x > 0, as you are aware, so the domain for the derivative is x > 0, as well.

As it turns out, the function y = 1/x is defined for any $x \ne 0$, but the left-hand branch does not represent the derivative of the natural log function.

3. Jan 13, 2016

### lightarrow

I assume you are referring to real numbers (you use "x" for the ind. variable and you say that lnx is defined for positive x only) because in the complex field it's all another story...

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lightarrow

4. Jan 13, 2016

### mathman

ln(x) can be defined for x < 0, using $x=-xe^{\pi i}$. Therefore $ln(x)=ln(-x)+\pi i$.

5. Jan 14, 2016

### Svein

In fact, $\frac{d}{dx}ln(\lvert x \rvert)=\frac{1}{x}$.