🎯 Logarithmic Differentiation (Advanced Technique)

What is Logarithmic Differentiation?

Logarithmic differentiation is a powerful technique that uses properties of logarithms to make complicated derivatives much easier!

When to Use This Technique

Use logarithmic differentiation when you have:

1. Products of many functions: $(x+1)(x+2)(x+3)$
2. Quotients with complicated parts: $\frac{\sqrt{x}(x+1)^3}{(x-2)^4}$
3. Variable in both base and exponent: $x^x$, $(x^2)^{\sin x}$, $x^{\ln x}$
4. Messy combinations: Functions that would require Product Rule, Quotient Rule, and Chain Rule all at once

💡 Key Idea: Take ln of both sides, simplify using log properties, then differentiate!

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The General Process

Step-by-Step Method

1. Take ln of both sides: $\ln y = \ln[f(x)]$
2. Simplify using log properties:
   • $\ln(ab) = \ln a + \ln b$
   • $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$
   • $\ln(a^n) = n\ln a$
3. Differentiate implicitly with respect to $x$
4. Solve for $\frac{dy}{dx}$: Multiply both sides by $y$
5. Substitute back: Replace $y$ with the original function

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Example 1: Variable Base and Exponent

Find $\frac{dy}{dx}$ if $y = x^x$

This is impossible with our usual rules! But logarithmic differentiation makes it easy.

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Step 1: Take ln of both sides

$\ln y = \ln(x^x)$

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Step 2: Use log property $\ln(a^n) = n\ln a$

$\ln y = x\ln x$

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Step 3: Differentiate both sides implicitly

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

Right side (Product Rule): $\frac{d}{dx}[x\ln x] = (1)\ln x + x\cdot\frac{1}{x} = \ln x + 1$

So: $\frac{1}{y}\frac{dy}{dx} = \ln x + 1$

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Step 4: Solve for $\frac{dy}{dx}$

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

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Step 5: Substitute $y = x^x$

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

Answer: $\frac{dy}{dx} = x^x(\ln x + 1)$

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Example 2: Complicated Products and Quotients

Find $\frac{dy}{dx}$ if $y = \frac{x^3\sqrt{x+1}}{(2x-1)^4}$

Without logarithmic differentiation, this would require Quotient Rule, Product Rule, and Chain Rule!

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Step 1: Take ln of both sides

$\ln y = \ln\left[\frac{x^3\sqrt{x+1}}{(2x-1)^4}\right]$

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Step 2: Simplify using log properties

$\ln y = \ln(x^3) + \ln\sqrt{x+1} - \ln[(2x-1)^4]$

$\ln y = 3\ln x + \frac{1}{2}\ln(x+1) - 4\ln(2x-1)$

Now it's just a sum/difference instead of a messy quotient!

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Step 3: Differentiate both sides

$\frac{1}{y}\frac{dy}{dx} = 3\cdot\frac{1}{x} + \frac{1}{2}\cdot\frac{1}{x+1} - 4\cdot\frac{2}{2x-1}$

$\frac{1}{y}\frac{dy}{dx} = \frac{3}{x} + \frac{1}{2(x+1)} - \frac{8}{2x-1}$

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Step 4: Multiply by $y$

$\frac{dy}{dx} = y\left[\frac{3}{x} + \frac{1}{2(x+1)} - \frac{8}{2x-1}\right]$

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Step 5: Substitute original function

$\frac{dy}{dx} = \frac{x^3\sqrt{x+1}}{(2x-1)^4}\left[\frac{3}{x} + \frac{1}{2(x+1)} - \frac{8}{2x-1}\right]$

Answer: $\displaystyle\frac{dy}{dx} = \frac{x^3\sqrt{x+1}}{(2x-1)^4}\left[\frac{3}{x} + \frac{1}{2(x+1)} - \frac{8}{2x-1}\right]$

💡 Note: This is much easier than using Quotient Rule + Product Rule!

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Why It Works

Logarithmic differentiation works because:

1. Log properties turn products into sums and quotients into differences
2. Sums and differences are much easier to differentiate than products and quotients
3. The Chain Rule for $\ln y$ gives us $\frac{1}{y}\frac{dy}{dx}$, which we can solve

It's essentially using algebra to simplify before calculus!

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Common Patterns

Pattern 1: Variable Exponent Only

$y = a^{f(x)}$ where $a$ is constant

Example: $y = 3^{x^2}$

$\ln y = \ln(3^{x^2}) = x^2\ln 3$

$\frac{1}{y}\frac{dy}{dx} = 2x\ln 3$

$\frac{dy}{dx} = 2x(\ln 3) \cdot 3^{x^2}$

Pattern 2: Variable Base Only

$y = [g(x)]^n$ where $n$ is constant

Example: $y = (x^2 + 1)^{10}$

This can also be done with Chain Rule, but log differentiation works:

$\ln y = 10\ln(x^2 + 1)$

$\frac{1}{y}\frac{dy}{dx} = 10 \cdot \frac{2x}{x^2+1}$

$\frac{dy}{dx} = \frac{20x}{x^2+1} \cdot (x^2+1)^{10} = 20x(x^2+1)^9$

Pattern 3: Both Variable

$y = [g(x)]^{h(x)}$

Example: $y = (\sin x)^x$

$\ln y = x\ln(\sin x)$

$\frac{1}{y}\frac{dy}{dx} = \ln(\sin x) + x\cdot\frac{\cos x}{\sin x}$

$\frac{dy}{dx} = (\sin x)^x[\ln(\sin x) + x\cot x]$

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Advantages Over Other Methods

Product Rule vs. Logarithmic Differentiation

Product of 3 functions: $(x+1)(x+2)(x+3)$

Product Rule (messy!):

• Takes 3 applications
• Many terms to combine

Logarithmic Differentiation (clean!): $\ln y = \ln(x+1) + \ln(x+2) + \ln(x+3)$ $\frac{1}{y}\frac{dy}{dx} = \frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3}$

Much simpler! ✓

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⚠️ Common Mistakes

Mistake 1: Forgetting to Multiply by y

After differentiating, you have $\frac{1}{y}\frac{dy}{dx} = ...$

Don't forget to multiply both sides by $y$!

Mistake 2: Wrong Log Properties

❌ $\ln(x + y) = \ln x + \ln y$ (WRONG!) ✅ $\ln(xy) = \ln x + \ln y$ (multiplication becomes addition)

Mistake 3: Not Substituting Back

After solving for $\frac{dy}{dx}$, remember to substitute the original expression for $y$!

Mistake 4: Domain Issues

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

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When NOT to Use Logarithmic Differentiation

Don't use it for simple functions where standard rules work fine:

• Simple powers: $x^5$ — just use Power Rule
• Simple products: $x\sin x$ — just use Product Rule
• Exponential base $e$: $e^{3x}$ — just use Chain Rule

Use logarithmic differentiation for complicated expressions where it genuinely simplifies your work!

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Special Case: Products of Many Terms

For $y = f_1(x) \cdot f_2(x) \cdot f_3(x) \cdots f_n(x)$:

$\ln y = \ln f_1 + \ln f_2 + \ln f_3 + \cdots + \ln f_n$

$\frac{1}{y}\frac{dy}{dx} = \frac{f_1'}{f_1} + \frac{f_2'}{f_2} + \frac{f_3'}{f_3} + \cdots + \frac{f_n'}{f_n}$

$\frac{dy}{dx} = y\left[\frac{f_1'}{f_1} + \frac{f_2'}{f_2} + \cdots + \frac{f_n'}{f_n}\right]$

💡 Formula: Each factor contributes $\frac{\text{derivative of factor}}{\text{factor}}$

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📝 Practice Strategy

1. Recognize when logarithmic differentiation will help (products, quotients, variable exponents)
2. Take ln of both sides
3. Simplify thoroughly using all log properties before differentiating
4. Differentiate implicitly (left side gives $\frac{1}{y}\frac{dy}{dx}$)
5. Multiply by $y$ to isolate $\frac{dy}{dx}$
6. Substitute the original expression for $y$
7. Simplify if possible (but complicated answers are okay!)