Differentiation of log(1+x)- Calculus Tutorial with Examples

How to Differentiate log(1+x)

Let's cut through the noise. The derivative of log(1+x) is one of the most useful tools you'll encounter in calculus. It shows up in growth models, probability, and signal processing. You need to know it cold.

The formula is straightforward:

d/dx [log(1+x)] = 1/(1+x)

That's it. But if you want to understand why this works and how to apply it, keep reading.

The Chain Rule Connection

This derivative comes from the chain rule. You're not differentiating a simple log(x). You're differentiating log(u) where u = 1+x.

The general rule is:

d/dx [log(u)] = (1/u) × du/dx

Plug in u = 1+x:

• du/dx = 1
• 1/u = 1/(1+x)
• Multiply: (1/(1+x)) × 1 = 1/(1+x)

Natural Log vs. Common Log

Quick clarification. When mathematicians write log(x) without a base, they mean the natural logarithm ln(x). This is what calculus textbooks use.

If you're working with log₁₀(x), the derivative changes:

d/dx [log₁₀(1+x)] = 1/((1+x) × ln(10))

Most calculus problems use the natural log version. Know which one you're dealing with before you start.

Solved Examples

Example 1: Basic Differentiation

Find d/dx [log(1+x)]

This is the direct application:

d/dx [log(1+x)] = 1/(1+x)

Domain restriction: x ≠ -1. The original function only exists for x > -1 anyway.

Example 2: With a Coefficient

Find d/dx [3log(1+x)]

Use the constant multiple rule. Constants pull out:

d/dx [3log(1+x)] = 3 × (1/(1+x)) = 3/(1+x)

Example 3: Chain Rule Inside and Out

Find d/dx [log(1+x²)]

Here u = 1+x², so du/dx = 2x:

d/dx [log(1+x²)] = (1/(1+x²)) × 2x = 2x/(1+x²)

Example 4: Composite Function

Find d/dx [sin(x) × log(1+cos(x))]

This requires the product rule:

Let u = sin(x) and v = log(1+cos(x))

v' = (1/(1+cos(x))) × (-sin(x)) = -sin(x)/(1+cos(x))

Answer: cos(x) × log(1+cos(x)) + sin(x) × [-sin(x)/(1+cos(x))]

You can simplify further using trig identities, but this shows the structure.

Quick Reference Table

Common Mistakes to Avoid

• Forgetting the chain rule. Always check what's inside the log. If it's not just x, you need to multiply by the derivative of the inside.
• Wrong base. Mixing up ln(x) and log₁₀(x) will give you the wrong answer. Confirm which one your problem uses.
• Domain errors. log(1+x) requires 1+x > 0, so x > -1. Don't ignore this in applied problems.
• Over-simplifying. Sometimes 1/(1+x) is the final answer. Don't force unnecessary algebra.

How to Get Started

Here's your step-by-step process for any log differentiation problem:

1. Identify the inner function. What's inside the log? Call it u.
2. Take du/dx. Differentiate the inner function.
3. Apply the formula. d/dx[log(u)] = (1/u) × du/dx
4. Substitute back. Replace u with the original expression.
5. Simplify. Cancel factors if possible.

Practice with three problems before you move on. Work through:

• d/dx [log(1+x³)]
• d/dx [2log(1+4x)]
• d/dx [log(1+ln(x))]

Check your answers against the pattern. If you're stuck on number 3, remember: the inside of the log is 1+ln(x), and the derivative of ln(x) is 1/x.

When This Shows Up in Real Problems

This derivative isn't just academic. You'll see it in:

• Logistic growth functions — modeling population with carrying capacity
• Information theory — entropy calculations use log functions
• Finance — continuously compounded returns involve ln(1+r)
• Signal processing — logarithmic amplifiers and filters

Understanding the mechanics here makes those applications much easier to handle.