Logarithmic Differentiation

Logarithmic differentiation is useful when the function to be differentiated is a product or quotient of multiple factors, or when it contains exponents that complicate direct differentiation.

Consider the function

\[ f(x) = \frac{\sqrt{x^2+1} (x+2)}{(x+1)^{1/3} x^2}. \]

Differentiating this directly would be cumbersome due to the presence of products, quotients, and exponents. To simplify the process, we take the natural logarithm on both sides:

\[ \ln f(x) = \ln \left( \frac{\sqrt{x^2+1} (x+2)}{(x+1)^{1/3} x^2} \right). \]

Using logarithmic properties,

\[ \ln f(x) = \frac{1}{2} \ln (x^2+1) + \ln (x+2) - \frac{1}{3} \ln (x+1) - 2 \ln x. \]

Now, differentiating both sides,

\[ \frac{1}{f(x)} \frac{d}{dx} f(x) = \frac{1}{2} \cdot \frac{2x}{x^2+1} + \frac{1}{x+2} - \frac{1}{3} \cdot \frac{1}{x+1} - \frac{2}{x}. \]

Simplifying,

\[ \frac{1}{f(x)} \frac{d}{dx} f(x) = \frac{x}{x^2+1} + \frac{1}{x+2} - \frac{1}{3(x+1)} - \frac{2}{x}. \]

Multiplying both sides by \( f(x) \),

\[ \frac{d}{dx} f(x) = f(x) \left( \frac{x}{x^2+1} + \frac{1}{x+2} - \frac{1}{3(x+1)} - \frac{2}{x} \right). \]

Substituting back \( f(x) \),

\[ \frac{d}{dx} f(x) = \frac{\sqrt{x^2+1} (x+2)}{(x+1)^{1/3} x^2} \left( \frac{x}{x^2+1} + \frac{1}{x+2} - \frac{1}{3(x+1)} - \frac{2}{x} \right). \]

Thus, logarithmic differentiation simplifies the problem by breaking the function into additive logarithmic terms before applying differentiation.

Example

Differentiate \( y = (x+1)(x+2)^2(x+3)^3(x+4)^4 \) using logarithmic differentiation.

Solution:

Taking the natural logarithm on both sides,

\[ \ln y = \ln \left( (x+1)(x+2)^2(x+3)^3(x+4)^4 \right). \]

Using logarithmic properties to expand,

\[ \ln y = \ln (x+1) + 2 \ln (x+2) + 3 \ln (x+3) + 4 \ln (x+4). \]

Differentiating both sides,

\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x+1} + 2 \cdot \frac{1}{x+2} + 3 \cdot \frac{1}{x+3} + 4 \cdot \frac{1}{x+4}. \]

Multiplying both sides by \( y \),

\[ \frac{dy}{dx} = y \left( \frac{1}{x+1} + \frac{2}{x+2} + \frac{3}{x+3} + \frac{4}{x+4} \right). \]

Substituting back \( y = (x+1)(x+2)^2(x+3)^3(x+4)^4 \),

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

Example

Differentiate \( y = \frac{x^y \sqrt{x+1}}{(x+2)(x^2 - x + 1)} \) using logarithmic differentiation.

Solution:

Taking the natural logarithm on both sides,

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

Using logarithmic properties,

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

Rewriting,

\[ \ln y = y \ln x + \frac{1}{2} \ln (x+1) - \ln (x+2) - \ln (x^2 - x + 1). \]

Differentiating both sides,

\[ \frac{1}{y} \frac{dy}{dx} = \ln x \frac{dy}{dx} + y \frac{1}{x} + \frac{1}{2(x+1)} - \frac{1}{x+2} - \frac{2x-1}{x^2 - x + 1}. \]

Rearranging,

\[ \frac{1}{y} \frac{dy}{dx} - \ln x \frac{dy}{dx} = y \frac{1}{x} + \frac{1}{2(x+1)} - \frac{1}{x+2} - \frac{2x-1}{x^2 - x + 1}. \]

Factoring \( \frac{dy}{dx} \),

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

Solving for \( \frac{dy}{dx} \),

\[ \frac{dy}{dx} = \frac{\frac{y}{x} + \frac{1}{2(x+1)} - \frac{1}{x+2} - \frac{2x-1}{x^2 - x + 1}}{\frac{1}{y} - \ln x}. \]