This formula only works for $x > 0$ since $\ln x$ is only defined for positive numbers.

💡 Connection: Since $e^x$ and $\ln x$ are inverse functions, their derivatives are reciprocals in a sense!

The Chain Rule with ln

When taking the derivative of $\ln$ of a function, use the Chain Rule:

General Formula

$\frac{d}{dx}[\ln u] = \frac{u'}{u} = \frac{1}{u} \cdot u'$

where $u$ is any function of $x$.

Examples

1. $\frac{d}{dx}[\ln(2x)] = \frac{2}{2x} = \frac{1}{x}$

2. $\frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2} = \frac{2}{x}$

3. $\frac{d}{dx}[\ln(3x + 1)] = \frac{3}{3x + 1}$

4. $\frac{d}{dx}[\ln(\sin x)] = \frac{\cos x}{\sin x} = \cot x$

💡 Pattern: Derivative of the inside over the inside!

Logarithms with Other Bases

For logarithms with base $a$:

$\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}$

Examples

1. $\frac{d}{dx}[\log_{10} x] = \frac{1}{x \ln 10}$

2. $\frac{d}{dx}[\log_2 x] = \frac{1}{x \ln 2}$

Why We Prefer ln

This is why calculus uses natural logs ($\ln$) instead of $\log_{10}$:

• Formula is simpler: just $\frac{1}{x}$
• No extra $\ln a$ factor to worry about!

Properties of Logarithms (For Simplifying)

Before differentiating, often use log properties to simplify:

Key Properties

1. $\ln(ab) = \ln a + \ln b$

2. $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$

3. $\ln(a^n) = n\ln a$

4. $\ln(e^x) = x$ and $e^{\ln x} = x$

Example Using Properties

Find $\frac{d}{dx}\left[\ln\left(\frac{x^2}{x+1}\right)\right]$

Method 1: Direct (harder) $\frac{d}{dx}\left[\ln\left(\frac{x^2}{x+1}\right)\right] = \frac{1}{\frac{x^2}{x+1}} \cdot \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2}$ (messy!)

Method 2: Simplify first (easier) $\ln\left(\frac{x^2}{x+1}\right) = \ln(x^2) - \ln(x+1) = 2\ln x - \ln(x+1)$

Now differentiate: $\frac{d}{dx}[2\ln x - \ln(x+1)] = \frac{2}{x} - \frac{1}{x+1}$

Much simpler! ✓

Logarithmic Differentiation

A powerful technique for complicated products and quotients:

When to Use

Use logarithmic differentiation when you have:

• Products of many functions
• Quotients with complicated numerators and denominators
• Variable bases and exponents: $x^x$, $x^{\sin x}$, etc.

The Process

1. Take $\ln$ of both sides
2. Use log properties to simplify
3. Differentiate implicitly
4. Solve for $\frac{dy}{dx}$
5. Substitute back to eliminate $y$

Example: $y = x^x$

Step 1: Take ln of both sides $\ln y = \ln(x^x) = x\ln x$

Step 2: Differentiate (implicitly on left, product rule on right) $\frac{1}{y} \cdot \frac{dy}{dx} = (1)\ln x + (x)\frac{1}{x} = \ln x + 1$

Step 3: Solve for $\frac{dy}{dx}$ $\frac{dy}{dx} = y(\ln x + 1)$

Step 4: Substitute $y = x^x$ $\frac{dy}{dx} = x^x(\ln x + 1)$

⚠️ Common Mistakes

Mistake 1: Forgetting the Chain Rule

❌ $\frac{d}{dx}[\ln(x^2)] = \frac{1}{x^2}$ ✅ $\frac{d}{dx}[\ln(x^2)] = \frac{2x}{x^2} = \frac{2}{x}$

Mistake 2: Wrong Simplification

Using $\ln(x^2) = 2\ln x$ is valid only when $x > 0$! Actually, $\ln(x^2) = 2\ln|x|$ to handle negative $x$.

Mistake 3: Mixing Up Rules

❌ $\frac{d}{dx}[\ln x] = \frac{1}{x^2}$ (confused with $\frac{d}{dx}\left[\frac{1}{x}\right]$) ✅ $\frac{d}{dx}[\ln x] = \frac{1}{x}$

Mistake 4: Domain Issues

$\ln x$ is only defined for $x > 0$. If needed for all $x \neq 0$, use $\ln|x|$.

Special Derivatives Involving ln

Derivative of Absolute Value

$\frac{d}{dx}[\ln|x|] = \frac{1}{x}$

This works for both positive and negative $x$ (but not $x = 0$).

Useful Combination

$\frac{d}{dx}[x\ln x] = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

Integration Connection

Since $\frac{d}{dx}[\ln x] = \frac{1}{x}$, we have:

$\int \frac{1}{x}\,dx = \ln|x| + C$

Applications

Exponential Growth/Decay

If $y = Ce^{kt}$, then $\ln y = \ln C + kt$

Differentiating: $\frac{1}{y}\frac{dy}{dt} = k$, so $\frac{dy}{dt} = ky$

Economics: Elasticity

Elasticity $E = \frac{p}{q} \cdot \frac{dq}{dp}$ can be written using logs:

$E = \frac{d(\ln q)}{d(\ln p)}$

Relative Rate of Change

$\frac{d}{dx}[\ln f(x)] = \frac{f'(x)}{f(x)}$ gives the relative (or percentage) rate of change.

📝 Key Formulas to Memorize

1. $\frac{d}{dx}[\ln x] = \frac{1}{x}$

2. $\frac{d}{dx}[\ln u] = \frac{u'}{u}$ (Chain Rule)

3. $\frac{d}{dx}[\log_a x] = \frac{1}{x\ln a}$

4. $\frac{d}{dx}[\ln|x|] = \frac{1}{x}$

Practice Tips

1. Simplify first using log properties when possible
2. Use logarithmic differentiation for products, quotients, and variable exponents
3. Remember $\frac{d}{dx}[\ln u] = \frac{u'}{u}$ - "derivative of inside over inside"
4. Factor when you get expressions like $\frac{2x}{x^2} = \frac{2}{x}$
5. Check domains - ln is only defined for positive arguments

Using $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$:

$y = \ln(x^2\sqrt{x+1}) - \ln(x-1)$

Using $\ln(ab) = \ln a + \ln b$:

$y = \ln(x^2) + \ln\sqrt{x+1} - \ln(x-1)$

Using $\ln(a^n) = n\ln a$ and $\sqrt{x+1} = (x+1)^{1/2}$:

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

Step 2: Now differentiate (much easier!)

$\frac{dy}{dx} = 2 \cdot \frac{1}{x} + \frac{1}{2} \cdot \frac{1}{x+1} - \frac{1}{x-1}$

$= \frac{2}{x} + \frac{1}{2(x+1)} - \frac{1}{x-1}$

Answer: $\displaystyle\frac{dy}{dx} = \frac{2}{x} + \frac{1}{2(x+1)} - \frac{1}{x-1}$

Key Lesson: Always simplify with log properties BEFORE differentiating!

$\ln y = x\ln(2x+1)$

Step 3: Differentiate both sides

Left side (implicit): $\frac{d}{dx}[\ln y] = \frac{1}{y}\frac{dy}{dx}$

Right side (product rule): $\frac{d}{dx}[x\ln(2x+1)] = (1)\ln(2x+1) + (x)\frac{2}{2x+1}$

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

Step 4: Set them equal

$\frac{1}{y}\frac{dy}{dx} = \ln(2x+1) + \frac{2x}{2x+1}$

Step 5: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx} = y\left[\ln(2x+1) + \frac{2x}{2x+1}\right]$

Step 6: Substitute back $y = (2x+1)^x$

$\frac{dy}{dx} = (2x+1)^x\left[\ln(2x+1) + \frac{2x}{2x+1}\right]$

Answer: $\displaystyle\frac{dy}{dx} = (2x+1)^x\left[\ln(2x+1) + \frac{2x}{2x+1}\right]$