### Theorem The [[Derivative of a function|derivative]] of the [[Natural Logarithm]] is the [[Reciprocal Function]] $\frac{d}{d x}(\ln x)=\frac{1}{x}, \quad x>0$ $\frac{d}{d x} \ln u=\frac{1}{u} \frac{d u}{d x}, \quad u>0$ #### Proof Knowing that the natural log is the [[Inverse Function]] of the [[𝑒^x - Exponential Function]], combined with the [[Derivative of the natural exponential]]: $\begin{aligned} y &=\ln x \\ e^{y} &=x \\ \frac{d}{d x}\left(e^{y}\right) &=\frac{d}{d x}(x) \\ e^{y} \frac{d y}{d x} &=1 \\ \frac{d y}{d x} &=\frac{1}{e^{y}}=\frac{1}{x} \end{aligned}$ Or using the [[Derivative of an inverse]] $ \begin{aligned}\left(f^{-1}\right)^{\prime}(x) &=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)} \\ &=\frac{1}{e^{f^{-1}(x)}} \\ &=\frac{1}{e^{\ln x}} \\ &=\frac{1}{x} \end{aligned} $