Answer
Use the fact that $\frac{d \ln x}{dx}=\frac{1}{x}$ , the change-of base theorem for logarithms, is
$$
\log_{a}x=\frac{\ln x}{\ln a}
$$
Find the derivative of each side, by the quotient rule , we have:
$$
\begin{aligned}
\frac{d \log _{a} x}{d x} &=\frac{\ln a \cdot \frac{d \ln x}{d x}-\ln x \cdot \frac{d \ln a}{d x}}{(\ln a)^{2}} \\
&=\frac{\ln a \cdot \frac{1}{x}-\ln x \cdot 0}{(\ln a)^{2}} \\
&=\frac{\frac{1}{x}}{\ln a} \\
&=\frac{1}{x \ln a}.
\end{aligned}
$$
Work Step by Step
Use the fact that $\frac{d \ln x}{dx}=\frac{1}{x}$ , the change-of base theorem for logarithms, is
$$
\log_{a}x=\frac{\ln x}{\ln a}
$$
Find the derivative of each side, by the quotient rule , we have:
$$
\begin{aligned}
\frac{d \log _{a} x}{d x} &=\frac{\ln a \cdot \frac{d \ln x}{d x}-\ln x \cdot \frac{d \ln a}{d x}}{(\ln a)^{2}} \\
&=\frac{\ln a \cdot \frac{1}{x}-\ln x \cdot 0}{(\ln a)^{2}} \\
&=\frac{\frac{1}{x}}{\ln a} \\
&=\frac{1}{x \ln a}.
\end{aligned}
$$
