Derivative of x to the x/Proof 3
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Theorem
- $\dfrac \d {\d x} x^x = x^x \paren {\ln x + 1}$
Proof
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 $a = 1$.
$\blacksquare$