Derivative of x to the x

Theorem

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

for $x > 0$.

Proof

Note that the Power Rule cannot be used because the index is not a constant.

Define $y$ as $x^x$.

As $x$ was stipulated to be positive, we can take the natural logarithm of both sides:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \ln y\)	\(=\)	\(\displaystyle \ln x^x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \ln x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Laws of Logarithms
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac 1 y \frac {\mathrm dy} {\mathrm dx}\)	\(=\)	\(\displaystyle 1 \cdot \ln x + x \cdot \frac 1 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Differentiating both sides WRT $x$, Chain Rule, Product Rule, Derivative of Identity and Natural Logarithm functions.
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm dy} {\mathrm dx}\)	\(=\)	\(\displaystyle x^x(\ln x + 1)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Multiply both sides by $y = x^x$

$\blacksquare$

Sources

• For a video presentation of the contents of this page, visit the Khan Academy.