Derivatives of Logarithmic Functions

The derivative of a logarithmic function starts with one rule: the derivative of ln x is $\frac{1}{x}$ , for $x \gt 0$ . For another base, $\frac{d}{dx}\log_a x = \frac{1}{x \ln a}$ . With an inside function u, the chain rule gives $\frac{d}{dx}\ln u = \frac{u'}{u}$ : the derivative of the inside divided by the inside.

Before this lesson: Derivatives of Exponential Functions, Derivatives of Inverse Functions and Properties of Logarithms

Why the derivative of ln x is 1/x

The natural log is the inverse of $e^x$ . If $y = \ln x$ , then $e^y = x$ . Differentiate both sides with respect to x, using implicit differentiation:

e^y \cdot \frac{dy}{dx} = 1 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}

You can see it on the graph. Near x = 0 the curve $y = \ln x$ is very steep, and as x grows it flattens out. At x = ½ the slope is 2, at x = 1 it is 1, and at x = 4 it is only ¼.

Graph of y = ln x for x from 0 to 6. Dashed tangent lines are drawn at (1/2, ln 1/2) with slope 2, at (1, 0) with slope 1, and at (4, ln 4) with slope 1/4. The curve gets flatter as x increases. — The slope of ln x at any x is 1/x, so the curve flattens as x grows.

Logarithmic derivative rules

Function	Derivative	With an inside function u
$\ln x$	$\frac{1}{x}$	$\frac{u'}{u}$
$\ln\|x\|$	$\frac{1}{x}$	$\frac{u'}{u}$
$\log_a x$	$\frac{1}{x\ln a}$	$\frac{u'}{u\ln a}$

The rule for $\log_a x$ follows from the change-of-base formula: $\log_a x = \frac{\ln x}{\ln a}$ , and $\frac{1}{\ln a}$ is just a constant.

Use log properties first. Rewriting a log of a product, quotient or power as a sum of simpler logs often makes the derivative much easier (see properties of logarithms).

Derivatives of logarithmic functions examples

Example 1: A sum with ln x

Differentiate $y = 3\ln x + x^2$ .

y' = 3 \cdot \frac{1}{x} + 2x

The derivative of ln x is 1/x; power rule on x².

y' = \frac{3}{x} + 2x

Multiply.

Answer: $y' = \frac{3}{x} + 2x$ .

Example 2: Chain rule with ln

Differentiate $y = \ln(x^2 + 1)$ .

u = x^2 + 1, \; u' = 2x

The inside function is x² + 1.

y' = \frac{u'}{u} = \frac{2x}{x^2 + 1}

Derivative of the inside over the inside (see chain rule).

Answer: $y' = \frac{2x}{x^2 + 1}$ .

Example 3: Product rule with ln x

Differentiate $y = x\ln x$ .

y' = (1)(\ln x) + (x)\left(\frac{1}{x}\right)

Product rule (see product rule).

y' = \ln x + 1

Simplify: x · 1/x = 1.

Answer: $y' = \ln x + 1$ .

Example 4: Use log properties first

Differentiate $y = \ln\left(x^3\sqrt{x + 1}\right)$ .

y = \ln\left(x^3\right) + \ln\left((x + 1)^{1/2}\right)

The log of a product is the sum of the logs.

y = 3\ln x + \frac{1}{2}\ln(x + 1)

Bring each exponent down in front (power property).

y' = \frac{3}{x} + \frac{1}{2} \cdot \frac{1}{x + 1}

Differentiate each term. The inside x + 1 has derivative 1.

y' = \frac{3}{x} + \frac{1}{2(x + 1)}

Multiply.

Answer: $y' = \frac{3}{x} + \frac{1}{2(x + 1)}$ . Without the log properties, you would need the chain rule and the product rule together.

Example 5: Word problem with a growing tree

A young tree’s height is modeled by $h(t) = 20\ln(t + 1)$ feet, t years after planting. How fast is it growing at t = 4 years? At t = 19 years?

h'(t) = 20 \cdot \frac{1}{t + 1} = \frac{20}{t + 1}

Chain rule: the inside t + 1 has derivative 1.

h'(4) = \frac{20}{5} = 4

Substitute t = 4.

h'(19) = \frac{20}{20} = 1

Substitute t = 19.

Answer: the tree grows about 4 feet per year at 4 years and about 1 foot per year at 19 years. A log model always keeps growing, but more and more slowly.

Common mistakes

Mistake	Example of the error	Fix
Forgetting the chain rule	$\frac{d}{dx}\ln(x^2 + 1) = \frac{1}{x^2 + 1}$	Multiply by u′: $\frac{2x}{x^2 + 1}$ .
Using 1/x for any log base	$\frac{d}{dx}\log_2 x = \frac{1}{x}$	Divide by ln a too: $\frac{1}{x\ln 2}$ .
Splitting a log of a sum	$\ln(x^2 + 1) = \ln x^2 + \ln 1$	Log properties work on products, quotients and powers, not sums.
Treating (ln x)² like ln(x²)	$\frac{d}{dx}(\ln x)^2 = \frac{2}{x}$	Chain rule: $2\ln x \cdot \frac{1}{x}$ .
Differentiating a log constant	$\frac{d}{dx}\ln 5 = \frac{1}{5}$	ln 5 is a constant, so its derivative is 0.

Related methods

The derivative of ln|x|

The function $\ln|x|$ is defined for negative x too, and its derivative is still $\frac{1}{x}$ . This version matters later, because it is why the antiderivative of $\frac{1}{x}$ is $\ln|x| + C$ .

Logarithmic differentiation

You can also take the natural log of both sides of an equation before you differentiate. That trick handles messy products and powers like $x^x$ , and it is the next lesson, Logarithmic Differentiation.

Practice problems

Differentiate $y = \ln(5x)$ .
Differentiate $y = \ln(x^2 - 3x)$ .
Differentiate $y = x^2\ln x$ .
Differentiate $y = (\ln x)^2$ .
Differentiate $y = \log_{10} x$ .
Differentiate $y = \ln\sqrt{x}$ .
Differentiate $y = \ln\left(\frac{x + 1}{x - 1}\right)$ .
Find the tangent line to $y = \ln x$ at x = 1.
Find where $y = \frac{\ln x}{x}$ has a horizontal tangent line.
Find the slope of $y = \log_3(x^2 + 2)$ at x = 1.

Show answers

y' = \frac{5}{5x}

Derivative of the inside over the inside.

y' = \frac{1}{x}

Simplify. (Or: ln 5x = ln 5 + ln x, and ln 5 is a constant.)

Answer: $y' = \frac{1}{x}$

y' = \frac{2x - 3}{x^2 - 3x}

The inside has derivative 2x − 3.

Answer: $y' = \frac{2x - 3}{x^2 - 3x}$

y' = 2x\ln x + x^2 \cdot \frac{1}{x}

Product rule.

y' = 2x\ln x + x

Simplify x²/x = x.

Answer: $y' = 2x\ln x + x$

y' = 2\ln x \cdot \frac{1}{x}

General power rule: the inside ln x has derivative 1/x.

y' = \frac{2\ln x}{x}

Multiply.

Answer: $y' = \frac{2\ln x}{x}$

y' = \frac{1}{x\ln 10}

Rule for log base a with a = 10.

Answer: $y' = \frac{1}{x\ln 10}$

y = \frac{1}{2}\ln x

Power property: √x = x^(1/2).

y' = \frac{1}{2x}

Differentiate.

Answer: $y' = \frac{1}{2x}$

y = \ln(x + 1) - \ln(x - 1)

Quotient property.

y' = \frac{1}{x + 1} - \frac{1}{x - 1}

Differentiate each term.

y' = \frac{(x - 1) - (x + 1)}{(x + 1)(x - 1)}

Use a common denominator (see adding and subtracting rational expressions).

y' = \frac{-2}{x^2 - 1}

Simplify the numerator and multiply out the denominator.

Answer: $y' = -\frac{2}{x^2 - 1}$

y(1) = \ln 1 = 0

Find the point (1, 0).

y'(1) = \frac{1}{1} = 1

The slope is 1/x at x = 1.

y - 0 = 1(x - 1)

Point-slope form (see point-slope form).

Answer: $y = x - 1$

y' = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2}

Quotient rule (see quotient rule).

y' = \frac{1 - \ln x}{x^2}

Simplify the numerator.

1 - \ln x = 0 \Rightarrow \ln x = 1

Set the numerator equal to 0.

x = e

Rewrite in exponential form.

Answer: x = e, at the point (e, 1/e)

10.

y' = \frac{2x}{(x^2 + 2)\ln 3}

Rule for log base 3 with the chain rule.

y'(1) = \frac{2}{3\ln 3}

Substitute x = 1.

\frac{2}{3\ln 3} \approx 0.607

Use ln 3 ≈ 1.0986.

Answer: about 0.607

Derivatives of logarithmic functions FAQ

What is the derivative of ln x?

$\frac{d}{dx}\ln x = \frac{1}{x}$ for $x \gt 0$ . The same formula holds for ln|x| when x is negative.

What is the derivative of ln(u)?

By the chain rule, $\frac{d}{dx}\ln u = \frac{u'}{u}$ . For example, the derivative of ln(3x + 2) is $\frac{3}{3x + 2}$ .

What is the derivative of log base a of x?

$\frac{d}{dx}\log_a x = \frac{1}{x\ln a}$ . For base 10, that is $\frac{1}{x\ln 10} \approx \frac{0.434}{x}$ .

Why is the derivative of ln(5x) the same as that of ln x?

Because ln 5x = ln 5 + ln x, and ln 5 is a constant with derivative 0. Multiplying the inside by a constant only shifts the log graph up or down.

Summary

$\frac{d}{dx}\ln x = \frac{1}{x}$ and $\frac{d}{dx}\log_a x = \frac{1}{x\ln a}$ .
Chain rule: $\frac{d}{dx}\ln u = \frac{u'}{u}$ .
Simplify with log properties before you differentiate.
Log properties don’t split a log of a sum.
The rule comes from ln x being the inverse of eˣ.

Previous lesson: Derivatives of Exponential Functions

Next lesson: Logarithmic Differentiation

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