Logarithmic Differentiation

When to Use

Logarithmic differentiation is useful when:

• The function has variable bases AND variable exponents: $f(x)^{g(x)}$
• The function is a complicated product or quotient
• You want to simplify before differentiating

The Method

1. Take the natural log of both sides: $\ln(y) = \ln(f(x))$
2. Use logarithm properties to simplify
3. Differentiate implicitly
4. Solve for $\frac{dy}{dx}$
5. Substitute back the original $y$

Logarithm Properties

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

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

$\ln(a^n) = n \cdot \ln(a)$

Examples

Example 1: Variable base and exponent

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

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

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

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

Example 2: General form f(x)^g(x)

Find $\frac{dy}{dx}$ for $y = (\sin x)^x$:

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

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

$= x \cot(x) + \ln(\sin x)$

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

Example 3: Complicated quotient

Find $\frac{dy}{dx}$ for $y = \frac{(x^2 + 1)^3(x - 1)^4}{(x + 2)^5}$:

$\ln(y) = 3\ln(x^2 + 1) + 4\ln(x - 1) - 5\ln(x + 2)$

$\frac{1}{y} \cdot \frac{dy}{dx} = 3 \cdot \frac{2x}{x^2 + 1} + 4 \cdot \frac{1}{x - 1} - 5 \cdot \frac{1}{x + 2}$

$= \frac{6x}{x^2 + 1} + \frac{4}{x - 1} - \frac{5}{x + 2}$

$\frac{dy}{dx} = y\left[\frac{6x}{x^2 + 1} + \frac{4}{x - 1} - \frac{5}{x + 2}\right]$

Example 4: x to a function power

Find $\frac{dy}{dx}$ for $y = x^{\sin x}$:

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

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

$= \frac{\sin(x)}{x} + \cos(x) \ln(x)$

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

Example 5: Tower of exponents

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

Let $u = x^x$, so $y = x^u$.

First find $\frac{du}{dx}$ (from Example 1):

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

Now for $y = x^u$:

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

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

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

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

Alternative Formula

For $y = f(x)^{g(x)}$:

$\frac{dy}{dx} = f(x)^{g(x)} \cdot \left[g'(x) \cdot \ln(f(x)) + g(x) \cdot \frac{f'(x)}{f(x)}\right]$

This comes directly from logarithmic differentiation.