1. Why Logarithmic Differentiation Is Useful
Logarithmic differentiation is helpful when a function is too complicated for direct differentiation. It simplifies expressions involving products, quotients, or powers, especially when the exponent itself contains variables. By taking logarithms, the structure becomes easier to differentiate.
2. Core Idea
The idea is simple: take the natural logarithm on both sides of the function, apply log rules to simplify, differentiate both sides, then solve for the derivative. This technique converts difficult expressions into manageable pieces.
3. Steps of Logarithmic Differentiation
- Take natural logarithm on both sides: ln(y).
- Use log identities to break down products, powers, and quotients.
- Differentiate both sides using implicit differentiation.
- Solve for dy/dx.
4. Useful Log Rules
- \( \ln(ab) = \ln a + \ln b \)
- \( \ln(a/b) = \ln a - \ln b \)
- \( \ln(a^n) = n \ln a \)
5. Examples
These examples show how the method simplifies expressions before differentiating.
5.1. Example 1 — Product Inside Power
Differentiate:
y = (x^2 + 1)^5
Take ln on both sides:
\ln y = 5 \ln(x^2 + 1)
Differentiate:
\frac{1}{y} \frac{dy}{dx} = 5 \cdot \frac{2x}{x^2 + 1}
Final derivative:
\frac{dy}{dx} = y \cdot \frac{10x}{x^2 + 1} = (x^2 + 1)^5 \cdot \frac{10x}{x^2 + 1}
5.2. Example 2 — Variable Exponent (x^x Type)
Differentiate:
y = x^x
Take ln:
\ln y = x \ln x
Differentiate:
\frac{1}{y} \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1
Final derivative:
\frac{dy}{dx} = x^x(\ln x + 1)
5.3. Example 3 — Complicated Product
Differentiate:
y = x^3 (3x + 2)^4 \sqrt{x - 1}
Take ln:
\ln y = 3\ln x + 4\ln(3x + 2) + \frac{1}{2}\ln(x - 1)
Differentiate each term and solve for dy/dx.
6. When Logarithmic Differentiation Helps Most
It is especially useful for:
- complicated products
- complicated quotients
- very large powers
- variable exponents like \( x^x \)
- expressions where direct rules become messy