Derivative of x to the x/Proof 1

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Theorem

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


Proof

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:

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

$\blacksquare$


Sources