Problem 38 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. $$ f(x)=\ln \frac{2 x}{1+x^{2}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Type

The given function is \( f(x) = \ln \left( \frac{2x}{1+x^2} \right) \), which is a logarithmic function. It involves the natural logarithm \( \ln(u) \) where \( u = \frac{2x}{1+x^2} \). We need to use the chain rule and the quotient rule here.

Step 2: Apply the Chain Rule

To differentiate \( f(x) \) with respect to \( x \), we first differentiate the outer function. If \( u(x) = \frac{2x}{1+x^2} \), then \( \ln(u) \) differentiates to \( \frac{1}{u} \cdot \frac{du}{dx} \). So, \( \frac{df}{dx} = \frac{1}{\frac{2x}{1+x^2}} \cdot \frac{d}{dx}\left( \frac{2x}{1+x^2} \right) \).

Step 3: Differentiate the Inner Function using the Quotient Rule

Use the quotient rule, which states \( \frac{d}{dx}\left( \frac{v}{w} \right) = \frac{v'w - vw'}{w^2} \), where \( v = 2x \) and \( w = 1 + x^2 \). Here, \( v' = 2 \) and \( w' = 2x \).

Step 4: Compute Derivative using the Quotient Rule

\( \frac{du}{dx} = \frac{(2)(1+x^2) - (2x)(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} \). Simplify this to get \( \frac{2 - 2x^2}{(1+x^2)^2} \).

Step 5: Insert \( \frac{du}{dx} \) back into the Differentiation Formula

Replace \( \frac{du}{dx} \) in the chain rule expression to get \( \frac{df}{dx} = \frac{1}{\frac{2x}{1+x^2}} \times \frac{2(1-x^2)}{(1+x^2)^2} \).

Step 6: Simplify the Expression

Simplify the expression by multiplying: \[ \frac{1+x^2}{2x} \cdot \frac{2(1-x^2)}{(1+x^2)^2} = \frac{2(1-x^2)}{2x(1+x^2)} = \frac{1-x^2}{x(1+x^2)} \]. This is the derivative \( f'(x) \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating complex functions. It applies when a function is composed of other functions, like peeling layers of an onion. If you have a function written as \( f(g(x)) \), we differentiate it by taking the derivative of the outer function \( f' \), leaving the inner unchanged, and then multiply this by the derivative of the inner function \( g'(x) \).

In our original exercise, \( f(x) = \ln(\frac{2x}{1+x^2}) \) uses a natural logarithm as the outer function and a quotient as the inner function \( u = \frac{2x}{1+x^2} \). To differentiate the natural log, treat it like this: if \( y = \ln(u) \), then its derivative \( \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \). This simple multiplication is the core of the chain rule.

This method allows you to break down the differentiation into manageable pieces. First, find the derivative of the outside function (here, the log part), then calculate the derivative of the inside (the quotient section), and multiply them together.

Quotient Rule

The quotient rule is specifically for situations when you have a function expressed as a fraction, or quotient, of two other functions. If you have \( v(x) \) over \( w(x) \), the quotient rule dictates the derivative formula:

• \( \frac{d}{dx}(\frac{v}{w}) = \frac{v'w - vw'}{w^2} \)

This formula highlights a critical point: we differentiate the numerator and denominator separately and unite them in this product/subtraction form.

For this equation: \( u(x) = \frac{2x}{1+x^2} \); the necessary derivatives are::

• The derivative of the numerator \( 2x \) is \( 2 \).
• The derivative of the denominator \( 1+x^2 \) is \( 2x \).

Plug these into the quotient rule to get: \[ \frac{(2)(1+x^2) - (2x)(2x)}{(1+x^2)^2} \].

After simplifying, this reduces to \( \frac{2 - 2x^2}{(1+x^2)^2} \). Without the quotient rule's organized approach, differentiating fractions would be much more complex.

Logarithmic Function

Logarithms, especially natural logs \( \ln(x) \), appear frequently in calculus. The logarithmic function helps in simplifying complex differentiation, especially when facing products or powers.

In our specific problem, the function begins as \( f(x)=\ln(\frac{2x}{1+x^2}) \), integrating a log function with a quotient. The property of \( \ln(u) \) dictates that its derivative becomes \( \frac{1}{u} \), reflecting the log rule formula:\

• \( \frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx} \)

This derivative rule allows you to handle logarithms almost like background work, dealing with the log's complex interior systematically.

In calculus, embracing logarithmic differentiation often makes otherwise baffling derivatives relatively straightforward. It provides a structured system where logs turn multiplication and division into manageable addition and subtraction tasks, easy to differentiate.