Derivative of Logarithmic Functions- Comprehensive Practice Problems

Derivatives of Logarithmic Functions: What You Actually Need to Know

Logarithmic derivatives show up everywhere in calculus. Physics, engineering, economics—they all use them. If you're still struggling with the basic rules, you're going to hit a wall fast.

This guide cuts through the noise. No philosophical digressions. Just the rules, the patterns, and practice problems that actually teach you something.

The Core Formula You Must Memorize

For the natural logarithm ln(x), the derivative is:

d/dx [ln(x)] = 1/x

That's it. One rule. But here's where it gets interesting—when you combine this with the chain rule, things get more complex.

Derivative of log_b(x) for Any Base

When your logarithm uses a base other than e, the formula changes:

d/dx [log_b(x)] = 1/(x · ln(b))

The ln(b) in the denominator is the only difference. If the base is e, then ln(e) = 1, and you're back to 1/x.

The Chain Rule: Where Most People Mess Up

Simple cases are easy. But what happens when the argument isn't just x?

d/dx [ln(u)] = u'/u

where u is a function of x.

This is the generalized form. You take the derivative of the inside, then divide by the inside. Practice this pattern until it becomes automatic.

Example:

d/dx [ln(3x² + 5)]

The inside is u = 3x² + 5, so u' = 6x.

Answer: 6x / (3x² + 5)

Comparing Derivative Rules

Function	Derivative	Condition
ln(x)	1/x	x > 0
log_b(x)	1/(x · ln(b))	x > 0, b > 0, b ≠ 1
ln(u)	u'/u	u > 0
ln\|f(x)\|	f'(x)/f(x)	f(x) ≠ 0

Practice Problems

Work through these. Don't peek at the solutions until you've tried them.

Problem 1

Find d/dx [ln(4x)]

Problem 2

Find d/dx [log₂(x³)]

Problem 3

Find d/dx [ln(x² + 3x + 2)]

Problem 4

Find d/dx [x · ln(x)]

Problem 5

Find d/dx [ln(sin(x))]

Problem 6

Find d/dx [log₁₀(5x² - 2)]

Problem 7

Find d/dx [(ln(x))²]

Problem 8

Find d/dx [eˣ · ln(eˣ + 1)]

Solutions

Solution 1: d/dx [ln(4x)]

Method 1: Use the logarithm property ln(4x) = ln(4) + ln(x)

Derivative of ln(4) is 0. Derivative of ln(x) is 1/x.

Answer: 1/x

Method 2: Use the chain rule directly.

Inside: u = 4x, so u' = 4

Derivative = 4/(4x) = 1/x

Solution 2: d/dx [log₂(x³)]

Using the base formula: d/dx [log_b(x³)] = (x³)' / (x³ · ln(2))

(x³)' = 3x²

Answer: 3x² / (x³ · ln(2)) = 3 / (x · ln(2))

Solution 3: d/dx [ln(x² + 3x + 2)]

Chain rule: u = x² + 3x + 2, u' = 2x + 3

Answer: (2x + 3) / (x² + 3x + 2)

Solution 4: d/dx [x · ln(x)]

This requires the product rule.

d/dx [x · ln(x)] = (1)(ln(x)) + (x)(1/x)

= ln(x) + 1

Solution 5: d/dx [ln(sin(x))]

Chain rule with u = sin(x), u' = cos(x)

Answer: cos(x) / sin(x) = cot(x)

Solution 6: d/dx [log₁₀(5x² - 2)]

Base 10 means ln(10) in the denominator.

u = 5x² - 2, u' = 10x

Answer: 10x / [(5x² - 2) · ln(10)]

Solution 7: d/dx [(ln(x))²]

Chain rule: treat this as [ln(x)]²

Derivative of u² is 2u · u'

u = ln(x), u' = 1/x

Answer: 2 · ln(x) · (1/x) = 2·ln(x)/x

Solution 8: d/dx [eˣ · ln(eˣ + 1)]

Product rule: d/dx [f·g] = f'·g + f·g'

f = eˣ, f' = eˣ

g = ln(eˣ + 1), g' = eˣ/(eˣ + 1)

Answer: eˣ · ln(eˣ + 1) + eˣ · [eˣ/(eˣ + 1)]

Simplified: eˣ · ln(eˣ + 1) + e^(2x)/(eˣ + 1)

Common Mistakes to Avoid

Forgetting to multiply by the derivative of the inside when using the chain rule
Confusing ln(x) derivative (1/x) with log base 10 derivative (1/(x·ln(10)))
Dropping the denominator entirely—this is the most common error
Forgetting the chain rule when the argument contains x

Getting Started: The Mental Checklist

When you see a logarithmic derivative problem, run through this:

Is the base e? If yes, derivative is 1/x (plus chain rule)
Is the base something else? Use 1/(x·ln(b))
Is the argument just x? You're done
Is the argument a function of x? Apply the chain rule: u'/u
Is it a product or composition? Use product/chain rule as needed

Master these patterns. The problems above cover the range of complexity you'll encounter. If you can solve all eight without help, you're ready for whatever comes next.