# The derivative $\frac{\mathrm d}{\mathrm dx} x^x=x^x\left(\ln x+1\right)$ is problematic for $x<0$

To take the derivative of $$x ^ x$$, we write

$$\dfrac {\mathrm d}{\mathrm dx} x^x=\dfrac {\mathrm d}{\mathrm dx} e^{\ln x^x}=\dfrac {\mathrm d}{\mathrm dx} e^{x\ln x}= e^{x\ln x}× \dfrac {\mathrm d}{\mathrm dx}(x\ln x)=x^x\left(\ln x+1\right)$$

Here is my problem:

If $$x\in\mathbb{Z^-}$$, then $$x^x\in\mathbb {R}$$. But, $$\ln x \not\in\mathbb {R}.$$

Because, $$\ln x$$ is defined only in the set of positive real numbers.

If, $$x \not\in\mathbb {Z^{-}}$$ and $$x\in\mathbb{R^{-}}$$, then $$x^x\in\mathbb {C}$$ and $$\ln x \in\mathbb {C}.$$

But, the problem occurs if $$x\in\mathbb{Z^-}.$$

So, $$x^x=e^{x\ln x}$$ doesn't hold for all real numbers. This makes the derivative result suspicious.

Where is the problem?

• Does this help? math.stackexchange.com/questions/1551470/domain-of-xx You can't draw the graph for $x\in\mathbb{R}^-$ since it isn't defined for all values of $x \in \mathbb{Q}^-$ Aug 24, 2020 at 7:20
• @safdar no. I wrote for $x\in\mathbb Z^-$
– user548054
Aug 24, 2020 at 8:30
• You cannot find differentiability for discrete points. so then it doesn't work. Aug 24, 2020 at 8:31
• Usually, $x^x$ is just undefined in the negatives.
– user65203
Aug 24, 2020 at 9:18

Not at all suspicious. You can't differentiate a function at an isolate point of the domain. So even if you extend the domain of $$x^x$$ to the negative integers, you cannot differentiate it at these points: how do you do the limit?

One could define $$x^x$$ for negative rational values of $$x$$ having odd denominators.

The set $$W=\{a/b: a,b\in\mathbb{Z}, a<0, b>0, b\text{ odd}\}$$ is even dense in $$(-\infty,0)$$, so it could be a good candidate for doing limits over it.

There is a problem, though: consider $$-1/3$$. In every neighborhood of $$-1/3$$ there are points $$x_0$$ in $$W$$ having even numerator and also points $$x_1$$ in $$W$$ having odd numerator. The value of $$x^x$$ at $$x_0$$ is positive, the value of $$x^x$$ at $$x_1$$ is negative. Therefore the function is not continuous at $$-1/3$$.

Hence, differentiability is out of question.

If you consider $$x^x$$ over the complex numbers, you have to choose a branch cut for the complex logarithm. Then the function is well defined and even analytic: $$x^x=\exp(x\log x)$$. Of course, in order to consider the derivative at $$-1$$ you need to do a cut different from the standard one that removes the negative $$x$$-semiaxis.

The differentiability of a function can only be found if it is continuous in an interval $$(a,b)$$.

$$x^x$$ is continuous only for $$x > 0$$. For $$x<0$$, the graph can only be drawn for some discrete points. Differentiability is not defined for this part of the graph.

$$\frac{\mathrm d (x^x)}{\mathrm{d}x}=x^x(\ln x+1)\quad \forall\quad x\in \mathbb{R}^+$$

• At the point $x=-2$, we have $f'(-2)=1/4 (1 + i π + \log(2))$
– user548054
Aug 24, 2020 at 8:52
• @Elementary If you are defining the function over some subset of the complex plane using the relation $z^z=e^{z\ln{(z)}}$ then the derivative you have provided is valid for any value within the domain. Aug 24, 2020 at 9:07
• @Elementary what subset are you defining this function over? $\mathbb{Z}^-$ or $\mathbb{C}$? Aug 24, 2020 at 9:09
• @PeterForeman how can we prove $z^z=e^{z\ln z}$ ?
– user548054
Aug 24, 2020 at 9:12
• @Elementary It's typically a definition rather than something to prove. How else do you suggest evaluating $\pi^\pi$? Aug 24, 2020 at 9:13