Derivatives of Logarithmic Functions

Function	Derivative	Condition
$\ln x$	${\displaystyle \frac{1}{x}}$
$\ln |x|$	${\displaystyle \frac{1}{x}}$	$x\ne 0$
${\log }_{a}x$	${\displaystyle \frac{1}{x\ln a}}$
$\ln u$ (chain)	${\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}$

Derivative of $\ln x$

Let $y=\ln x$. Using the four-step method:

Step 1: $y+\mathrm{\Delta }y=\ln (x+\mathrm{\Delta }x)$

Step 2: $\mathrm{\Delta }y=\ln (x+\mathrm{\Delta }x)-\ln x=\ln (1+{\displaystyle \frac{\mathrm{\Delta }x}{x}})$

Step 3: ${\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}={\displaystyle \frac{1}{x}}\ln {(1+{\displaystyle \frac{\mathrm{\Delta }x}{x}})}^{x/\mathrm{\Delta }x}$

Step 4: As $\mathrm{\Delta }x\to 0$, letting $u=\mathrm{\Delta }x/x\to 0$:

$$\frac{dy}{dx}=\frac{1}{x}\ln [\underset{u\to 0}{lim}(1+u{)}^{1/u}]=\frac{1}{x}\ln e=\frac{1}{x}$$\boxed{\frac{d}{dx}\ln x=\frac{1}{x},\qquad(x>0)\tag{a}}

Derivative of ${\log }_{a}x$

Write ${\log }_{a}x={\displaystyle \frac{\ln x}{\ln a}}$:

$$\frac{d}{dx}{\log }_{a}x=\frac{1}{\ln a}\frac{d}{dx}\ln x=\frac{1}{x\ln a}$$\boxed{\frac{d}{dx}\log_a x=\frac{1}{x\ln a},\qquad(x>0)\tag{b}}

Derivative of $\ln u$ via the Chain Rule

If is differentiable:

$$\frac{d}{dx}\ln u=\frac{1}{u}\frac{du}{dx}$$

When working with logarithms, first use the properties of logarithms to convert products to sums, quotients to differences, and powers to multiples, then differentiate.

Worked Examples

Logarithmic Differentiation

To differentiate $y=f(x)$ when $f$ is a product/quotient/power combination:

Derivatives of Real Powers of $x$

Using logarithmic differentiation, we can prove the Power Rule for all real exponents:

Let $y=x^r$ ($r$ any real number, ). Then $\ln |y|=r\ln |x|$. Differentiating:

\frac{y'}{y}=\frac{r}{x}\quad\Rightarrow\quad y'=r\frac{y}{x}=r\frac{x^r}{x}=rx^{r-1}

This confirms ${\displaystyle \frac{d}{dx}}{x}^r=r{x}^{r-1}$ for all real $r$.

Frequently Asked Questions

When should I use logarithmic differentiation?

Use it when the function involves products, quotients, and powers simultaneously, especially when the exponent is a variable. It converts multiplicative structure into additive structure, making differentiation much easier.

What is ${\displaystyle \frac{d}{dx}}\ln (x^2+1)$?

By the Chain Rule: ${\displaystyle \frac{1}{x^2+1}}\cdot 2x={\displaystyle \frac{2x}{x^2+1}}$.

Is $\ln |u|$ differentiable at $u=0$?

No, $\ln |u|$ is not defined (or differentiable) at $u=0$. The formula ${\displaystyle \frac{d}{dx}}\ln |u|={\displaystyle \frac{1}{u}}{\displaystyle \frac{du}{dx}}$ requires $u\ne 0$.