Problem 78 - Calculus for Biology and Medicine

Question

Differentiate $$ f(x)=\frac{a x^{2}}{k^{2}+x^{2}} $$ with respect to \(x\). Assume that \(a\) and \(k\) are positive constants.

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Function

The given function is a rational function of the form \( f(x) = \frac{u(x)}{v(x)} \), where \( u(x) = ax^2 \) and \( v(x) = k^2 + x^2 \). To differentiate it, we will apply the quotient rule.

Step 2: Apply the Quotient Rule

The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). We need to find \( u'(x) \) and \( v'(x) \).

Step 3: Differentiate the Numerator and Denominator

Differentiate \( u(x) = ax^2 \) to get \( u'(x) = 2ax \). Differentiate \( v(x) = k^2 + x^2 \) to get \( v'(x) = 2x \).

Step 4: Substitute into the Quotient Rule Formula

Substitute \( u'(x) = 2ax \), \( v(x) = k^2 + x^2 \), \( u(x) = ax^2 \), and \( v'(x) = 2x \) into the quotient rule formula: \[ f'(x) = \frac{(2ax)(k^2 + x^2) - (ax^2)(2x)}{(k^2 + x^2)^2} \]

Step 5: Simplify the Derivative Expression

Expand and simplify the expression in the numerator:\[(2ax)(k^2 + x^2) = 2axk^2 + 2ax^3\]\[(ax^2)(2x) = 2ax^3\]Substitute back and simplify:\[f'(x) = \frac{2axk^2 + 2ax^3 - 2ax^3}{(k^2 + x^2)^2} = \frac{2axk^2}{(k^2 + x^2)^2}\]

Step 6: Conclude with the Final Derivative

Thus, the derivative of the function \( f(x) = \frac{ax^2}{k^2 + x^2} \) is:\[ f'(x) = \frac{2axk^2}{(k^2 + x^2)^2} \].

Key Concepts

Rational Function

Rational functions are mathematical expressions that represent the ratio of two polynomial functions. These functions take the form \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials, and \( q(x) eq 0 \).
In the context of our problem, the rational function is \( f(x) = \frac{ax^2}{k^2 + x^2} \). Here, \( ax^2 \) is a simple polynomial in the numerator, and \( k^2 + x^2 \) serves as the polynomial in the denominator.
Rational functions are defined wherever the denominator is non-zero, which means they can become undefined if the denominator equals zero. This is crucial in calculus because points of discontinuity often affect the behavior of derivatives.
Understanding rational functions is fundamental, especially when dealing with their derivatives, because their characteristic structure directly influences how we apply differentiation rules, such as the quotient rule.

Quotient Rule

The quotient rule is a technique for finding the derivative of a function that is expressed as a quotient of two differentiable functions. It is particularly useful for rational functions and is defined as follows:

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

In this exercise, we identified our function as a quotient of two polynomials, with \( u(x) = ax^2 \) and \( v(x) = k^2 + x^2 \). This formula requires us to find the derivatives of both \( u(x) \) and \( v(x) \) before plugging them back into the quotient rule formula.
Differentiation of the numerator gives us \( u'(x) = 2ax \), and for the denominator, \( v'(x) = 2x \). By substituting these derivatives into the quotient rule, we derive the function's overall derivative. This method ensures that we handle the differentiation correctly, treating both components of the rational function with equal care.

Derivative

A derivative in calculus represents the rate at which a function is changing at any given point. More formally, it is the limit of the average rate of change of the function over an infinitesimally small interval. For the rational function \( f(x) = \frac{ax^2}{k^2 + x^2} \), the derivative tells us how \( f(x) \) changes as \( x \) changes.
Through applying the quotient rule, we successfully determine the expression for the derivative of the function. In this problem, once simplified, the derivative emerges as \( f'(x) = \frac{2axk^2}{(k^2 + x^2)^2} \).
This expression provides a comprehensive insight into how the function behaves, indicating areas where the function increases or decreases, and identifying any peculiarities such as horizontal tangents or turning points. Understanding and calculating derivatives are essential in analyzing and predicting the behavior of mathematical models in fields such as physics, engineering, and beyond.