# 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

- 1953: L. Harwood Clarke:
*A Note Book in Pure Mathematics*... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Variable Index: Example $1$. - For a video presentation of the contents of this page, visit the Khan Academy.