Problem 38 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \frac{2 x}{1+x^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{1-x^2}{x(1+x^2)^2} \).

Step 1: Recognize the Need for Chain Rule

The function provided is \( f(x) = \ln \left(\frac{2x}{1+x^2}\right) \). This is a composite function with a natural logarithm acting on a fraction. We will need to apply the chain rule of differentiation to find the derivative.

Step 2: Identify the Inner Function

Inside the natural logarithm, our expression is \( \frac{2x}{1+x^2} \). This is what we will call our inner function, \( u(x) = \frac{2x}{1+x^2} \). We will differentiate \( u(x) \) in the next step.

Step 3: Differentiate the Inner Function

To find the derivative of \( u(x) = \frac{2x}{1+x^2} \), apply the quotient rule: \ \( u'(x) = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} \). \ Simplifying gives: \( u'(x) = \frac{2 + 0 - 4x^2}{(1+x^2)^2} = \frac{2 - 4x^2}{(1+x^2)^2} \).

Step 4: Differentiate the Logarithm Function

The derivative of \( \ln(u(x)) \), where \( u(x) \) is the inner function, is \( \frac{1}{u(x)} \times u'(x) \). Hence, \( f'(x) = \frac{1}{\frac{2x}{1+x^2}} \times \frac{2 - 4x^2}{(1+x^2)^2} \).

Step 5: Simplify the Derivative Expression

Calculate \( f'(x) = \frac{1+x^2}{2x} \times \frac{2 - 4x^2}{(1+x^2)^2} \). Simplifying this, we have: \ \( f'(x) = \frac{(2 - 4x^2)(1+x^2)}{2x(1+x^2)^2} \).

Step 6: Final Simplification

Expanding the numerator, we get \( 2(1 + x^2) - 4x^2(1 + x^2) \). This simplifies to \( 2 + 2x^2 - 4x^2 - 4x^4 \) which becomes \( 2 - 2x^2 - 4x^4 \). Divide each term by \( 2x(1+x^2)^2 \) to complete the simplification: \( f'(x) = \frac{2(1 - x^2)}{2x(1+x^2)^2} \). Finally, this simplifies to \( f'(x) = \frac{1-x^2}{x(1+x^2)^2} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composition of two or more functions. It allows us to differentiate complex expressions by breaking them down into simpler parts. When you have a function like \( f(g(x)) \), where \( f \) is a function of \( g(x) \), the chain rule helps in differentiating it. The rule states:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

So, you differentiate the outer function evaluated at the inner function and multiply it by the derivative of the inner function. This process makes it easier to handle nested functions, where you deal with each layer one at a time. In our exercise, the natural logarithm is the outer function, and the quotient is the inner function, making it a classic case for applying the chain rule.

Quotient Rule

The quotient rule is a handy tool for finding derivatives of functions that are ratios, meaning one function divided by another. To use the quotient rule, remember the formula:

• If \( y = \frac{u(x)}{v(x)} \), then the derivative \( y' \) is given by:\[ y' = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]

This means you take the derivative of the top function \( u(x) \), multiply it by the bottom function \( v(x) \), and then subtract the derivative of the bottom function \( v'(x) \) multiplied by the top function \( u(x) \). Finally, divide the entire expression by the square of the bottom function \( v(x) \). In our specific problem, when we differentiated the inner function \( \frac{2x}{1+x^2} \), we used this rule to find its derivative. The quotient rule simplifies the process of handling equations that involve fractions, ensuring you don’t miss any critical terms during differentiation.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions, and they appear frequently in calculus, especially when dealing with growth and decay problems. The logarithmic function with base \( b \) is defined as \( \, \log_b(x) \), but in calculus, we often use the natural logarithm, denoted by \( \, \ln(x) \), which has a base of \( e \). When differentiating natural logarithms:

• For \( \, \ln(x) \), the derivative is \( \, \frac{1}{x} \).
• If there is a more complex expression inside the logarithm, such as \( \, \ln(u(x)) \), you apply the chain rule and multiply \( \, \frac{1}{u(x)} \) by \( \, u'(x) \).

This principle was applied when differentiating \( \ln \left( \frac{2x}{1+x^2} \right) \) in our problem. Ensuring you understand how to differentiate these correctly is vital for solving a broad range of calculus problems involving exponential growth or decay, where the models are often expressed using logarithms.

Natural Logarithm

The natural logarithm, represented by \( \, \ln(x) \), is a fundamental function in calculus working with a constant base \( e \), a key number approximately equal to 2.718. This type of logarithm is particularly useful for differentiating and integrating functions that involve continuous growth or decay processes, as \( e \) appears naturally in many calculus formulas. To differentiate a natural logarithm:

• For a simple function like \( \, \ln(x) \), the derivative is \( \, \frac{1}{x} \).
• For more complex expressions, applying the chain rule allows us to take derivatives of \( \, \ln(u(x)) \) by first finding \( \, u'(x) \) and then multiplying it by \( \, \frac{1}{u(x)} \).

In our specific case, the function \( \ln \left( \frac{2x}{1+x^2} \right) \) required the application of the chain rule once we found the derivative of the inner function. The natural logarithm plays a significant role in many real-world applications like finance and physics, where continuous change is modeled.