$\begin{align}  & y={{\log }_{e}}x \\ & y+\delta y={{\log }_{e}}(x+\delta x) \\ & \delta y={{\log }_{e}}(x+\delta x)-{{\log }_{e}}x \\ & \delta y={{\log }_{e}}\left( \frac{x+\delta x}{x} \right)={{\log }_{e}}\left( 1+\frac{\delta x}{x} \right) \\ & \text{Divide through by }\delta x \\ & \frac{\delta y}{\delta x}=\frac{1}{\delta x}{{\log }_{e}}\left( 1+\frac{\delta x}{x} \right)=\frac{1}{x}\times \frac{x}{\delta x}{{\log }_{e}}\left( 1+\frac{\delta x}{x} \right) \\ & \frac{\delta y}{\delta x}=\frac{1}{x}{{\log }_{e}}{{\left( 1+\frac{\delta x}{x} \right)}^{\tfrac{x}{\delta x}}} \\ & \text{Let }\tfrac{x}{\delta x}=p \\ & \frac{\delta y}{\delta x}=\frac{1}{x}{{\log }_{e}}{{\left( 1+\frac{1}{p} \right)}^{p}} \\ & \text{Proceeding to their limits as }\delta x\to 0,p\to \infty ,{{\left( 1+\frac{1}{p} \right)}^{p}}\to e \\ & \frac{dy}{dx}=\underset{\delta x\to \infty }{\mathop{\lim }}\,\frac{1}{x}{{\log }_{e}}{{\left( 1+\frac{1}{p} \right)}^{p}}=\frac{1}{x}{{\log }_{e}}e \\ & \frac{dy}{dx}=\frac{1}{x} \\ & \therefore \text{Hence, }\frac{d}{dx}({{\log }_{e}}x)=\frac{1}{x} \\\end{align}$