Differentiating logarithmic functions

The derivative of ln x, the standard form for ln f(x), using log laws to simplify before differentiating, and the product and quotient rules with logarithms.

Standard forms

Logarithmic derivatives

$$\frac{d}{dx}\ln x = \frac1x \qquad \frac{d}{dx}\ln(ax + b) = \frac{a}{ax + b} \qquad \frac{d}{dx}\ln f(x) = \frac{f'(x)}{f(x)}$$

The gradient of y = ln x

The gradient of the tangent at x = P is 1/P: steep near the y-axis, and flatter and flatter to the right.

Try this

Find the point where the tangent has gradient $\tfrac12$.

Example 1●

Differentiate (a) $\ln(3x + 2)$ and (b) $\ln(x^2 + 1)$.

$\dfrac{3}{3x + 2}$

$\dfrac{2x}{x^2 + 1}$

Simplify first with the log laws

Example 2●●

Differentiate (a) $\ln 5x$ and (b) $\ln\dfrac{x + 1}{x - 1}$.

$\ln 5x = \ln 5 + \ln x$, so the derivative is $\dfrac1x$.

$\ln(x + 1) - \ln(x - 1)$, so the derivative is $\dfrac{1}{x + 1} - \dfrac{1}{x - 1} = \dfrac{-2}{x^2 - 1}$.

Product and quotient rules

Example 3●●

Differentiate (a) $x\ln x$ and (b) $\dfrac{\ln x}{x}$.

$\ln x + x \times \dfrac1x = \ln x + 1$

$\dfrac{x \times \frac1x - \ln x}{x^2} = \dfrac{1 - \ln x}{x^2}$

Flashcards

$dfrac{d}{dx}ln x$

$\dfrac1x$

$dfrac{d}{dx}ln f(x)$

$\dfrac{f'(x)}{f(x)}$

$dfrac{d}{dx}ln kx$

$\dfrac1x$, because $\ln kx = \ln k + \ln x$.

Practice

$\dfrac{d}{dx}\ln(3x + 2) = $
$\dfrac{d}{dx}\ln(x^2 + 1) = $
$\dfrac{d}{dx}\ln 5x = $
$\dfrac{d}{dx}x\ln x = $
$\dfrac{d}{dx}\dfrac{\ln x}{x} = $
$\dfrac{d}{dx}\ln\dfrac{x + 1}{x - 1} = $
Find the gradient of $y = \ln x$ at $x = 4$.
Find the gradient of $y = \ln(2x + 1)$ at $x = 0$.
Find the gradient of $y = \ln x^3$ at $x = 3$.