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 $y = x^{a x}$ |
$\blacksquare$