Problem 32 - Calculus for Biology and Medicine

Question

Differentiate $$ f(N)=\frac{b N^{2}+N}{K+b} $$

Step-by-Step Solution

Answer

The derivative is \( f'(N) = \frac{2bNK + 2b^2N + K + b}{(K + b)^2} \).

Step 1: Identify Differentiation Rules

The given function is \( f(N) = \frac{b N^2 + N}{K + b} \). This is a rational function, which will be differentiated using the quotient rule. The quotient rule states that if you have a function \( g(x) = \frac{u(x)}{v(x)} \), then its derivative \( g'(x) \) is given by \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Identify \( u(x) = bN^2 + N \) and \( v(x) = K + b \).

Step 2: Calculate the Derivatives of the Numerator and Denominator

Differentiate the numerator \( u(N) = bN^2 + N \). Use the power rule: \( u'(N) = 2bN + 1 \).Differentiate the denominator \( v(N) = K + b \). Since \( K + b \) is a constant, \( v'(N) = 0 \).

Step 3: Apply the Quotient Rule

Plug these derivatives into the quotient rule formula. Specifically: \( f'(N) = \frac{(2bN + 1)(K + b) - (bN^2 + N)(0)}{(K + b)^2} \). Since the derivative of the denominator \( v'(N) = 0 \), the expression simplifies to: \( f'(N) = \frac{(2bN + 1)(K + b)}{(K + b)^2} \).

Step 4: Simplify the Derivative

Expand the numerator: \( (2bN + 1)(K + b) = 2bNK + 2b^2N + K + b \). So, \( f'(N) = \frac{2bNK + 2b^2N + K + b}{(K + b)^2} \). This is the derivative of \( f(N) \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding how a function changes as its input changes. It tells us the rate at which something is changing at any given point. In simple terms, if you imagine a curve on a graph, differentiating that curve will give you the slope of the tangent line at any point on the curve.

When dealing with functions like polynomials or more complex functions, we use different rules and techniques to differentiate them correctly. In this exercise, we are looking at a rational function where you'll need to apply specific rules such as the quotient rule. Understanding the nature of the function is key to selecting the right method for differentiation.

Additionally, think of differentiation as a tool that converts a function describing your starting position and velocity to one that describes acceleration. It breaks down something potentially quite complex into simpler pieces, thus providing insight into the behavior and characteristics of functions.

Quotient Rule

The quotient rule is a differentiation technique used to find the derivative of a function that is the division of two differentiable functions. Symbolically, if you have a function in the form \( g(x) = \frac{u(x)}{v(x)} \), the derivative is given by:
\[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]
This rule emerges from the need to handle fractions or rational functions effectively when you're finding derivatives. It provides a structured way to differentiate without having to resort to ineffectual trial and error.

For example, with the function \( f(N) = \frac{bN^2 + N}{K + b} \), we identify:

• \( u(N) = bN^2 + N \)
• \( v(N) = K + b \)

Differentiating these gives us \( u'(N) = 2bN + 1 \) and \( v'(N) = 0 \), since \( K + b \) is a constant. Understanding when and how to apply the quotient rule is crucial in tackling derivatives of rational functions.

Power Rule

The power rule is one of the simplest and most widely used rules in differentiation. It helps us to differentiate functions of the form \( x^n \), where \( n \) is any real number. According to this rule, if \( y = x^n \), then the derivative \( y' = n x^{n-1} \).

This rule is particularly handy with polynomial functions and significantly simplifies the process of finding derivatives. In the exercise, the power rule is applied to the numerator \( bN^2 + N \) to obtain \( u'(N) \). Specifically, the terms \( bN^2 \) and \( N \) are differentiated to get \( 2bN \) and \( 1 \), respectively.

The power rule reduces complexity by allowing us to directly calculate the derivative term by term. Always remember, the power rule applies independently to each term of the polynomial, making it an incredibly efficient technique for differentiation.