Problem 37 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. $$ f(x)=\ln \frac{x}{x+1} $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = \ln \frac{x}{x+1} \) is \( \frac{1}{x(x+1)} \).

Step 1: Express the Function Using Logarithmic Properties

The given function is \( f(x) = \ln \frac{x}{x+1} \). We can simplify this expression using the logarithm property \( \ln \frac{a}{b} = \ln a - \ln b \). Thus:\[ f(x) = \ln x - \ln(x+1) \].

Step 2: Differentiate the Logarithmic Functions Separately

Now we need to take the derivative of each logarithmic function separately. The derivative of \( \ln u \) with respect to \( x \) is \( \frac{1}{u} \frac{du}{dx} \).For \( \ln x \), the derivative is:\[ \frac{d}{dx}(\ln x) = \frac{1}{x} \].For \( \ln(x+1) \), let \( u = x+1 \). Then:\[ \frac{d}{dx}(\ln(x+1)) = \frac{1}{x+1} \frac{d}{dx}(x+1) = \frac{1}{x+1}\].

Step 3: Subtract the Derivatives

Combine the derivatives from the previous step. According to the subtraction rule of differentiation, we have:\[ \frac{d}{dx}(f(x)) = \frac{d}{dx}(\ln x) - \frac{d}{dx}(\ln(x+1)) = \frac{1}{x} - \frac{1}{x+1} \].

Step 4: Simplify the Resulting Expression

The final step is to simplify the expression \( \frac{1}{x} - \frac{1}{x+1} \). Use the common denominator approach:\[ \frac{d}{dx}(f(x)) = \frac{x+1 - x}{x(x+1)} = \frac{1}{x(x+1)} \].

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is an invaluable technique for differentiating functions that involve logarithms. This technique hinges on a fundamental logarithm property:

• The property \( \ln \frac{a}{b} = \ln a - \ln b \) helps transform complex trigonometric fractions into simpler subtracted forms.
• By rewriting complex functions using logarithmic properties, we simplify the differentiation process.

In this exercise, the function \( f(x) = \ln \frac{x}{x+1} \) was transformed into two separate functions: \( f(x) = \ln x - \ln(x+1) \). This step paves the way for manageable derivative calculation, as each logarithmic component can be tackled individually. Understanding this initial breakdown is crucial because it sets the stage for applying other differentiation rules effectively.

Chain Rule

The chain rule is a critical concept in calculus for finding derivatives of composite functions. When you have a function within another function, the chain rule allows you to differentiate them correctly. The mathematical notation is commonly expressed as:\[ \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \]In this exercise:

• We determined the derivative of \( \ln(x+1) \) by treating \( x+1 \) as another function, \( u \), inside a logarithmic function.
• This step required recognizing the inner function \( u = x+1 \), allowing us to use the rule softly by taking \( \frac{du}{dx} \).

The chain rule made it straightforward to determine: \[ \frac{d}{dx}(\ln(x+1)) = \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) = \frac{1}{x+1} \]Leveraging the chain rule simplifies differentiating nested functions and ensures accuracy. This approach significantly reduces the complexity of more involved mathematical operations.

Simplifying Expressions

After differentiation, it is often necessary to simplify resulting expressions to achieve a cleaner, more interpretable form. Simplifying derivatives involves basic algebraic manipulation, including:

• Combining fractions for similar denominators.
• Reducing terms using elementary algebraic operations.

In this problem, the result after differentiating was: \[ \frac{1}{x} - \frac{1}{x+1} \]Upon simplifying:

• Adopt a common denominator: \( x(x+1) \).
• Rearrange terms: \[ \frac{x+1 - x}{x(x+1)} = \frac{1}{x(x+1)} \]

The simplification process not only clarifies the derivative but often reveals more about the underlying relationships within the original function, which is essential for deeper understanding or further applications.