### Theorem
$
\frac{d}{d x} a^{x}=a^{x} \ln a \qquad a>0
$
$
\frac{d}{d x} a^{u}=a^{u} \ln a \frac{d u}{d x} \qquad a>0
$
#### Proof
Rewrite using base [[π - Euler's Number|π]], and apply the [[Derivative of the natural exponential]] with [[Derivative Chain Rule|chain rule]], and then simplify back to base $a$
$
\begin{aligned} \frac{d}{d x} a^{x} &=\frac{d}{d x} e^{x \ln a} \\ &=e^{x \ln a} \cdot \frac{d}{d x}(x \ln a) \\ &=a^{x} \ln a \end{aligned}
$