Is ln(x) differentiable at negative x-axis

[Miraj Kayastha]

Since lnx is defined for positive x only shouldn't 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?

 

[Mark44]

Since lnx is defined for positive x only shouldn't 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?

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.

 

[lightarrow]

Since lnx is defined for positive x only shouldn't 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?

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

 

[mathman]

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

 

[Svein]

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