Differentiating Logarithmic Functions
Learn differentiating logarithmic functions for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-04). The natural logarithm \(\ln x\) has the surprisingly simple derivative \(\dfrac{1}{x}\) for positive \(x\), and this result can be shown from the fact that \(e\) and \(\ln\) undo each other.
You will learn to differentiate \(\ln x\) and expressions such as \(\ln(3x + 1)\) with the chain rule, giving the derivative of the inside over the inside, and use log laws to simplify first — a time-saver throughout the calculus that follows.
Theory
The natural logarithm differentiates to \(\dfrac{1}{x}\). This Year 12 Mathematics Advanced topic (MAV-12-04) differentiates \(\ln x\) and \(\ln f(x)\) using the chain rule and log laws.
\(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\). With the chain rule, \(\dfrac{d}{dx}\ln f(x)=\dfrac{f'(x)}{f(x)}\) — the derivative of the inside over the inside.
Log laws often simplify first: \(\ln(3x)=\ln3+\ln x\), so its derivative is just \(\dfrac{1}{x}\). The domain of \(\ln x\) is \(x>0\).
Method
- Simplify with log laws if possible.
- Identify the inside \(f(x)\) and its derivative.
- Write \(\dfrac{f'(x)}{f(x)}\).
| \(\ln(3x)\) | \(=\) | \(\ln3+\ln x\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{1}{x}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{2x}{x^{2}+1}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{5}{5x-2}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{2}{x}\) |
| \(\text{at }x=1\) | \(=\) | \(2\) |
Common pitfalls
Frequently asked questions
What is the derivative of ln x?
It is 1 over x.
How do you differentiate ln of a function?
Use the chain rule: the derivative of ln f of x is f prime of x over f of x, the derivative of the inside divided by the inside.
What is the derivative of ln 3x?
Using log laws ln 3x equals ln 3 plus ln x, so the derivative is just 1 over x.
Why is the domain of ln x only positive numbers?
Because you can only take the logarithm of a positive number, so ln x is defined only for x greater than 0.