# Derivative of x to the x

## Theorem

Let $x \in \R$ be a real variable whose domain is the set of (strictly) positive real numbers $\R_{>0}$.

Then:

$\dfrac \d {\d x} x^x = x^x \paren {\ln x + 1}$

## Proof 1

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

Let $y := x^x$.

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

 $\displaystyle \ln y$ $=$ $\displaystyle \ln x^x$ $\displaystyle$ $=$ $\displaystyle x \ln x$ Laws of Logarithms $\displaystyle \map {\frac \d {\d x} } {\ln y}$ $=$ $\displaystyle \map {\frac \d {\d x} } {x \ln x}$ differentiating both sides with respect to $x$ $\displaystyle \frac 1 y \frac {\d y} {\d x}$ $=$ $\displaystyle \map {\frac \d {\d x} } {x \ln x}$ Chain Rule for Derivatives and Derivative of Natural Logarithm Function $\displaystyle$ $=$ $\displaystyle \map {\frac \d {\d x} } x \cdot \ln x + x \frac {\d} {\d x} \ln x$ Product Rule for Derivatives $\displaystyle$ $=$ $\displaystyle 1 \cdot \ln x + x \cdot \frac 1 x$ Derivative of Identity Function and Derivative of Natural Logarithm Function $\displaystyle$ $=$ $\displaystyle \ln x + 1$ simplification $\displaystyle \frac {\d y} {\d x}$ $=$ $\displaystyle x^x \paren {\ln x + 1}$ multiplying both sides by $y = x^x$

$\blacksquare$

## Proof 2

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

 $\displaystyle \frac \d {\d x} x^x$ $=$ $\displaystyle \frac \d {\d x} \map \exp {x \ln x}$ Definition 1 of Power (Algebra) $\displaystyle$ $=$ $\displaystyle \paren {\frac \d {\map \d {x \ln x} } \map \exp {x \ln x} } \paren {\map {\frac \d {\d x} } {x \ln x} }$ Chain Rule for Derivatives $\displaystyle$ $=$ $\displaystyle \map \exp {x \ln x} \paren {\map {\frac \d {\d x} } {x \ln x} }$ Derivative of Exponential Function $\displaystyle$ $=$ $\displaystyle \map \exp {x \ln x} \paren {\map {\frac \d {\d x} } x \cdot \ln x + x \frac \d {\d x} \ln x}$ Product Rule for Derivatives $\displaystyle$ $=$ $\displaystyle \map \exp {x \ln x} \paren {1 \cdot \ln x + x \cdot \frac 1 x}$ Derivative of Identity Function and Derivative of Natural Logarithm Function $\displaystyle$ $=$ $\displaystyle \map \exp {x \ln x} \paren {\ln x + 1}$ Real Multiplication Identity is One and Inverses for Real Multiplication $\displaystyle$ $=$ $\displaystyle x^x \paren {\ln x + 1}$ Definition 1 of Power (Algebra)

$\blacksquare$

## Proof 3

From Derivative of $x^{a x}$ we have:

$\dfrac \d {\d x} x^{a x} = a x^{a x} \paren {\ln x + 1}$

The result follows on setting $a = 1$.

$\blacksquare$