Problem 40 - Calculus for Biology and Medicine

Question

Differentiate $$ g(N)=r N(a-N)\left(1-\frac{N}{K}\right) $$

Step-by-Step Solution

Answer

The derivative is \( g'(N) = ra - \frac{raN}{K} - 2rN + \frac{2rN^2}{K} - \frac{rN(a-N)}{K} \).

Step 1: Apply the Product Rule

The function given is a product of multiple terms: \( g(N) = rN(a-N)\left(1-\frac{N}{K}\right) \). To differentiate this, we use the product rule which states \((uv)' = u'v + uv'\). In our case, we will split the function into two parts: \( u(N) = rN(a-N) \) and \( v(N) = \left(1-\frac{N}{K}\right) \). We first need to differentiate \( u(N) \) and \( v(N) \) with respect to \( N \).

Step 2: Differentiate u(N)

We have \( u(N) = rN(a-N) \). To differentiate, apply the product rule to the terms \( rN \) and \( (a-N) \). This gives: \( u'(N) = r(a-N) + rN(-1) = ra - 2rN \).

Step 3: Differentiate v(N)

For \( v(N) = 1-\frac{N}{K} \), differentiate with respect to \( N \). The derivative is \( v'(N) = -\frac{1}{K} \).

Step 4: Apply Product Rule to g(N)

Now apply the product rule: \( g'(N) = u'(N)v(N) + u(N)v'(N) \). Substitute \( u'(N) = ra - 2rN \), \( v(N) = 1-\frac{N}{K} \), \( u(N) = rN(a-N) \), and \( v'(N) = -\frac{1}{K} \) into the product rule equation.

Step 5: Simplify the Expression

Substitute into the product rule: \( g'(N) = (ra - 2rN)\left(1-\frac{N}{K}\right) + rN(a-N)(-\frac{1}{K}) \). Expand and simplify each term: \( g'(N) = (ra - 2rN)\left(1-\frac{N}{K}\right) = ra - \frac{raN}{K} - 2rN + \frac{2rN^2}{K} + (-\frac{rN(a-N)}{K}) \). Simplify to obtain the derivative: \( g'(N) = ra - \frac{raN}{K} - 2rN + \frac{2rN^2}{K} - \frac{rN(a-N)}{K} \).

Key Concepts

Product Rule

The product rule is a fundamental technique used in calculus when you want to find the derivative of a product of two functions. In simpler terms, if you have two functions multiplied together, like \( u(N) \) and \( v(N) \), you can find the derivative of the product \( (uv)' \) by using this formula:

• \( (uv)' = u'v + uv' \)

Here, \( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \) respectively. In our given function \( g(N) = rN(a-N)\left(1-\frac{N}{K}\right) \), we split the expression into two parts. Let's say \( u(N) = rN(a-N) \) and \( v(N) = \left(1-\frac{N}{K}\right) \).Before applying the product rule, we first need to differentiate each part separately. Then, we use the product rule formula to combine them back to get the derivative of the entire expression. It might sound complex at first glance, but with practice, using the product rule becomes quite straightforward.

Derivative

A derivative represents the rate at which a function is changing at any given point. It's like the speed of a car; it tells you how fast the car’s position is changing over time. In calculus, finding the derivative of a function provides critical insights into the behavior and characteristics of the function.For example, in our exercise, we're interested in how \( g(N) = rN(a-N)\left(1-\frac{N}{K}\right) \) changes with respect to \( N \). The derivative \( g'(N) \) tells us this rate of change. Calculating derivatives typically involves rules and techniques suited for different function types, like the product rule for multiplying functions. Finding and understanding derivatives is vital in various fields like physics, economics, and statistics, because they help us predict and optimize outcomes.

Chain Rule

The chain rule is another essential tool we use when dealing with composite functions, though it might not explicitly appear in our problem's step-by-step solution, it's worth understanding how it works. This rule is crucial when a function is nested within another function.In a typical scenario where the chain rule is applied, you have something like \( h(x) = f(g(x)) \). To differentiate \( h(x) \), we use the chain rule:

• \( h'(x) = f'(g(x)) \cdot g'(x) \)

This rule allows us to differentiate the outside function \( f \) as if \( g(x) \) were a simple variable, and then multiply by the derivative of the inside function \( g(x) \).In the context of our problem, we primarily used the product rule, but you might encounter more complex exercises where the chain rule is essential for finding derivatives of more complicated expressions.

Simplification

Simplification is the final step after obtaining a derivative, where the expression is cleaned up to make it more understandable and easier to work with. Initially, after applying the product rule, the derivative can look messy, filled with expanded terms like our example.For \( g'(N) = (ra - 2rN)\left(1-\frac{N}{K}\right) + rN(a-N)(-\frac{1}{K}) \), simplifying involves expanding the terms and then combining like terms to reduce redundancy:

• First, distribute terms within each part.
• Then, combine similar terms wherever possible.

This helps in reducing the complexity of the equation, presenting a more "user-friendly" version of your results. Simplified expressions are especially helpful when they are used for further mathematical operations or analysis.In our problem, simplification results in a clearer understanding of how \( g(N) \) behaves as \( N \) changes, providing an efficient equation that is easier to interpret or calculate with.