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Derivative of General Logarithms

Reference: Stewart 6.4 • Chapter: 6 • Section: 4

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Why Base $e$ is Special

You already know that $\frac{d}{dx}(\ln x) = \frac{1}{x}$. What about logarithms with other bases, like $\log_{10} x$ or $\log_2 x$?

Here's the key insight: every logarithm can be written in terms of natural log. The change of base formula gives us:

$$\log_a x = \frac{\ln x}{\ln a}$$

Since $\ln a$ is just a constant, differentiating $\log_a x$ is nearly identical to differentiating $\ln x$; we just pick up an extra constant factor of $\frac{1}{\ln a}$.

This explains why mathematicians prefer natural logarithms in calculus: base $e$ is the only base where that extra factor equals 1.

Prerequisite Map

Before You Start

Quick check: Do you know the change of base formula?

Test: Write $\log_2 x$ in terms of natural logarithms.

Answer: $\log_2 x = \frac{\ln x}{\ln 2}$

If this is unfamiliar, recall: for any bases $a$ and $b$, $$\log_a x = \frac{\log_b x}{\log_b a}$$

Quick Reference

Formula	Domain
$\displaystyle\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$	$x > 0$, $a > 0$, $a \neq 1$
$\displaystyle\frac{d}{dx}[\log_a(g(x))] = \frac{g'(x)}{g(x) \ln a}$	$g(x) > 0$

Special cases:

$\displaystyle\frac{d}{dx}(\log_{10} x) = \frac{1}{x \ln 10} \approx \frac{0.434}{x}$
$\displaystyle\frac{d}{dx}(\log_2 x) = \frac{1}{x \ln 2} \approx \frac{1.443}{x}$
$\displaystyle\frac{d}{dx}(\ln x) = \frac{1}{x}$ (since $\ln e = 1$)

Key Concepts

The Main Formula and Its Derivation

$$\boxed{\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}}$$

Derivation:

Step 1. Use the change of base formula: $$\log_a x = \frac{\ln x}{\ln a}$$

Step 2. Since $\ln a$ is a constant, factor it out: $$\frac{d}{dx}(\log_a x) = \frac{d}{dx}\left(\frac{\ln x}{\ln a}\right) = \frac{1}{\ln a} \cdot \frac{d}{dx}(\ln x)$$

Step 3. Apply the known derivative of $\ln x$: $$= \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}$$

Why $e$ is special: When $a = e$, we have $\ln e = 1$, so the formula simplifies to $\frac{1}{x}$. For any other base, there's an extra factor of $\frac{1}{\ln a}$.

The Chain Rule Form

When the argument is a function $g(x)$ instead of just $x$:

$$\boxed{\frac{d}{dx}[\log_a(g(x))] = \frac{g'(x)}{g(x) \ln a}}$$

The pattern: derivative of inside over (inside times $\ln a$).

Important Notes on Domain

The function $\log_a x$ requires:

$x > 0$ (the input must be positive)
$a > 0$ (the base must be positive)
$a \neq 1$ (the base cannot be 1, since $\log_1 x$ is undefined)

Practice Problems

Level 1 Direct Application

Find $\frac{d}{dx}(\log_5 x)$.

Thought Process

Apply the formula directly with $a = 5$: $$\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$$

Show Answer

$$\frac{d}{dx}(\log_5 x) = \frac{1}{x \ln 5}$$

This can also be written as $\frac{1}{(\ln 5)x}$ or approximately $\frac{0.621}{x}$.

Level 2 Common Logarithm with Chain Rule

Differentiate $f(x) = \log_{10}(x^2 + 1)$.

Thought Process

Use the chain rule form with $g(x) = x^2 + 1$ and $a = 10$: $$\frac{d}{dx}[\log_a(g(x))] = \frac{g'(x)}{g(x) \ln a}$$

Compute $g'(x) = 2x$.

Show Answer

$$f'(x) = \frac{2x}{(x^2 + 1) \ln 10}$$

Note: $\ln 10 \approx 2.303$, so this is approximately $\frac{2x}{2.303(x^2 + 1)} \approx \frac{0.869x}{x^2 + 1}$.

Level 2 Trigonometric Argument

Find $\frac{d}{dx}[\log_3(4 - \cos x)]$.

Thought Process

Here $g(x) = 4 - \cos x$ and $a = 3$.

Find $g'(x) = \sin x$ (note the positive sign since derivative of $-\cos x$ is $\sin x$).

Apply the chain rule formula: $\frac{g'(x)}{g(x) \ln a}$.

Domain check: Since $-1 \leq \cos x \leq 1$, we have $3 \leq 4 - \cos x \leq 5$, so the argument is always positive. The domain is all real numbers.

Show Answer

$$\frac{d}{dx}[\log_3(4 - \cos x)] = \frac{\sin x}{(4 - \cos x) \ln 3}$$

Verification: At $x = 0$: $g(0) = 4 - 1 = 3$, $g'(0) = \sin 0 = 0$, so derivative is $0$. This makes sense since $\cos x$ has a maximum at $x = 0$, so $4 - \cos x$ has a minimum there, and the log function has a horizontal tangent at that point.

Level 3 Product Rule Combined

Find $\frac{d}{dx}[x \cdot \log_2 x]$.

Thought Process

This is a product of $x$ and $\log_2 x$.

Use the product rule: $(uv)' = u'v + uv'$

Let $u = x$ and $v = \log_2 x$.

$u' = 1$
$v' = \frac{1}{x \ln 2}$

Show Answer

Using the product rule: $$\frac{d}{dx}[x \cdot \log_2 x] = 1 \cdot \log_2 x + x \cdot \frac{1}{x \ln 2}$$

$$= \log_2 x + \frac{1}{\ln 2}$$

This can also be written as $\log_2 x + \log_2 e$ (since $\frac{1}{\ln 2} = \log_2 e$), which equals $\log_2(ex)$.

Level 4 Nested Logarithms

Differentiate $y = \log_2(x \log_3 x)$, where $x > 1$.

Thought Process

This is $\log_2$ of the expression $x \log_3 x$.

Let $g(x) = x \log_3 x$. We need $g'(x)$ first.

Use the product rule on $g(x)$:

$g'(x) = \log_3 x + x \cdot \frac{1}{x \ln 3} = \log_3 x + \frac{1}{\ln 3}$

Then apply the chain rule for the outer $\log_2$.

Show Answer

First, find the derivative of the inside: $$g(x) = x \log_3 x$$ $$g'(x) = \log_3 x + \frac{1}{\ln 3}$$

Now apply the chain rule: $$\frac{dy}{dx} = \frac{g'(x)}{g(x) \ln 2} = \frac{\log_3 x + \frac{1}{\ln 3}}{(x \log_3 x) \ln 2}$$

This can be simplified, but the form above is acceptable.

Level 5 Conceptual: Why Does the Base Matter?

Explain why the derivative of $\log_a x$ is smaller when $a > e$ and larger when $1 < a < e$.

Specifically: for $x = 10$, compare $\frac{d}{dx}(\log_2 x)$, $\frac{d}{dx}(\ln x)$, and $\frac{d}{dx}(\log_{10} x)$ evaluated at $x = 10$.

Thought Process

The derivative is $\frac{1}{x \ln a}$.

At fixed $x = 10$, the derivative is $\frac{1}{10 \ln a}$.

Consider how $\ln a$ varies with $a$:

When $a = 2$: $\ln 2 \approx 0.693$
When $a = e$: $\ln e = 1$
When $a = 10$: $\ln 10 \approx 2.303$

Since $\ln a$ is in the denominator, larger $\ln a$ means smaller derivative.

Show Answer

At $x = 10$:

Function	Derivative at $x=10$	Value
$\log_2 x$	$\frac{1}{10 \ln 2}$	$\approx 0.144$
$\ln x$	$\frac{1}{10}$	$= 0.100$
$\log_{10} x$	$\frac{1}{10 \ln 10}$	$\approx 0.043$

Explanation: The derivative $\frac{1}{x \ln a}$ is inversely proportional to $\ln a$.

When $a < e$: $\ln a < 1$, so $\frac{1}{\ln a} > 1$, making the derivative larger than $\frac{1}{x}$.
When $a = e$: $\ln a = 1$, so the derivative equals exactly $\frac{1}{x}$.
When $a > e$: $\ln a > 1$, so $\frac{1}{\ln a} < 1$, making the derivative smaller than $\frac{1}{x}$.

Geometric interpretation: $\log_2 x$ grows faster than $\ln x$, which grows faster than $\log_{10} x$. The derivative captures this: steeper growth means larger derivative.

Common Mistakes

Mistake	Why It's Wrong	Correct Approach
$\frac{d}{dx}(\log_{10} x) = \frac{1}{x}$	Forgot the $\ln a$ factor	$\frac{d}{dx}(\log_{10} x) = \frac{1}{x \ln 10}$
Writing $\log x$ without specifying base	Ambiguous (could be base 10 or base $e$)	Always write $\log_{10} x$ or $\ln x$ explicitly
$\frac{d}{dx}(\log_a x^2) = \frac{2x}{x^2 \ln a}$ is final	Not wrong, but can simplify	$= \frac{2}{x \ln a}$ (the $x$ cancels)

Mastery Checklist

[ ] I can state the formula $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$
[ ] I can derive this formula using change of base
[ ] I can apply the chain rule version for $\log_a(g(x))$
[ ] I understand why base $e$ gives the simplest derivative
[ ] I can combine this with product and quotient rules

Mental Model

Converting to natural log first:

When facing any $\log_a$ derivative, you can always convert:

Write $\log_a x = \frac{\ln x}{\ln a}$
Differentiate: $\frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}$

Or just memorize the formula and remember: it's like $\frac{1}{x}$, but with an extra $\ln a$ in the denominator.

Connections

Looking back:

This builds on Derivative of Natural Log
Uses the change of base formula from algebra

Looking ahead:

Logarithmic differentiation uses these ideas for complex functions
In applications, $\log_{10}$ appears in decibels and pH; $\log_2$ appears in computer science

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Derivative of Natural Log	Section 6.4	Integrals Yielding ln

Last updated: 2026-01-23