Derivative of x to the a x

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Theorem

Let $x \in \R$ be a real variable whose domain is the set of (strictly) positive real numbers $\R_{>0}$.

Let $c \in \R_{>0}$ be a (strictly) positive real constant.

Then:

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


Proof

Let $y := x^{a x}$.

As $x > 0$, we can take the natural logarithm of both sides:

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

$\blacksquare$