For $a>0$ and $a \neq 1$, $ \frac{d}{d x} \log _{a} u=\frac{1}{u \ln a} \frac{d u}{d x} $ ### Proof Rewrite using [[Change of Base]] $\begin{aligned} \frac{d}{d x} \log _{a} x &=\frac{d}{d x}\left(\frac{\ln x}{\ln a}\right) \\ &=\frac{1}{\ln a} \cdot \frac{d}{d x} \ln x \\ &=\frac{1}{\ln a} \cdot \frac{1}{x} \\ &=\frac{1}{x \ln a} \end{aligned}$