d/dx ln x – Sam Artigliere's Blog

Statement

$\frac{d}{dx}\ln x = \frac{1}{x}$

Proof

This proof is taken from Khan Academy:

https://www.khanacademy.org/math/in-in-grade-11-ncert/in-in-class11-differentiation/copy-of-sine-and-cosine-derivatives-ab/v/proof-of-derivative-of-ln-x

$\begin{array}{rcl} \frac{d}{dx}\left[ \ln x \right] &=& \displaystyle\lim_{\Delta x \to 0} \frac{\ln (x+\Delta x) - \ln x}{\Delta x}\\ \, &\,& \, \\ \ln a - \ln b &=& \ln\frac ab\quad\text{Therefore,}\\ \, &\,& \, \\ \frac{d}{dx}\left[ \ln x \right] &=& \displaystyle\lim_{\Delta x \to 0} \frac{\ln(\frac{x+\Delta x}{x})}{\Delta x}\\ \, &\,& \, \\ &=& \displaystyle\lim_{\Delta x \to 0} \frac{\ln(1+\frac{\Delta x}{x})}{\Delta x}\\ \, &\,& \, \\ a\ln b &=& \ln(b^a) \quad\text{thus,}\\ \, &\,& \, \\ \frac{d}{dx}\left[ \ln x \right] &=& \displaystyle\lim_{\Delta x \to 0}\frac{1}{\Delta x}\ln(1+\frac{\Delta x}{x})\\ \, &\,& \, \\ &=& \displaystyle\lim_{\Delta x \to 0} \ln\left[ (1+\frac{\Delta x}{x})^{\frac{1}{\Delta x}} \right]\\ \, &\,& \, \\ \text{Let }n&=&\frac{\Delta x}{x}\text\\ \, &\,& \, \\ \Delta x &=& nx \\ \, &\,& \, \\ \frac{1}{\Delta x}&=&\frac{1}{nx}=\frac{1}{n}\cdot\frac{1}{x}\\ \, &\,& \, \\ \text{as }\Delta x \to 0 &,& n \to 0 \quad\text{therefore, }\\ \, &\,& \, \\ \frac{d}{dx}\left[ \ln x \right] &=& \displaystyle\lim_{\Delta x \to 0} \ln\left[ (1+n)^{\frac{1}{n}\frac{1}{x}} \right]\\ \, &\,& \, \\ \frac{d}{dx}\left[ \ln x \right] &=& \displaystyle\lim_{\Delta x \to 0} \ln\left[ ((1+n)^{\frac{1}{n})\frac{1}{x}} \right]\\ \, &\,& \, \\ &=&\frac{1}{x}\displaystyle\lim_{\Delta x \to 0} \ln\left[ (1+n)^{\frac{1}{n} \right]\\ \, &\,& \, \\ &=&\frac{1}{x}\ln\displaystyle\lim_{\Delta x \to 0} \left[ (1+n)^{\frac{1}{n} \right]\\ \end{array}$

$\text{But }\displaystyle\lim_{\Delta x \to 0} \left[ (1+n)^{\frac{1}{n} \right] &=& e$ . Therefore,

$\frac{d}{dx}\left[ \ln x \right] &=& \frac{1}{x}\ln e = 1$

Thus,

$\frac{d}{dx}\left[ \ln x \right] &=& \frac{1}{x}\cdot 1 = \frac{1}{x}$