# Derivative of x to the x/Proof 1

## 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 $y = x^x$

$\blacksquare$