Derivative of x to the x

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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 $a = 1$.

$\blacksquare$