Problem 37 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is \( 1 - \frac{2N}{K} \).

Step 1: Apply Product Rule

The given function is a product of two functions: \( u = N \) and \( v = 1 - \frac{N}{K} \). Use the product rule for differentiation: \( \frac{d}{dN}(uv) = u'v + uv' \). First, differentiate \( u = N \), so \( u' = 1 \).

Step 2: Differentiate the Second Function

Now, differentiate the second function \( v = 1 - \frac{N}{K} \). This gives \( v' = -\frac{1}{K} \).

Step 3: Apply Product Rule Components

Substitute into the product rule: \( \frac{d}{dN}(N(1-\frac{N}{K})) = 1\cdot(1-\frac{N}{K}) + N\cdot(-\frac{1}{K}) \).

Step 4: Simplify the Expression

Simplify the expression from step 3: \( 1 - \frac{N}{K} - \frac{N}{K} = 1 - 2\frac{N}{K} \).

Step 5: Final Result

The derivative of \( g(N) \) is \( 1 - \frac{2N}{K} \).

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus for differentiating products of two functions. It states that to find the derivative of a product of two functions, you use the formula:\[\frac{d}{dx}(uv) = u'v + uv',\]where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives. This rule is crucial because it allows us to handle more complex expressions that can't be simplified into a single function.
For example, in our exercise with the function \( g(N) = N\left(1-\frac{N}{K}\right) \), the function is clearly a product of \( u = N \) and \( v = 1 - \frac{N}{K} \). The product rule lets us differentiate it effectively by finding the derivatives of each part and then combining them. In practice:

• Differentiating \( u = N \) gives \( u' = 1 \).
• Differentiating \( v = 1 - \frac{N}{K} \) gives \( v' = -\frac{1}{K} \).
• Apply the rule: \( \frac{d}{dN}(N(1-\frac{N}{K})) = 1\cdot(1-\frac{N}{K}) + N\cdot(-\frac{1}{K}) \).

Differentiation Steps

Differentiation involves a series of logical steps to find the rate at which a function changes. By breaking down problems systematically, as seen in our step-by-step solution, you can effectively tackle complex expressions. Here’s how it works:
First, identify the parts of the function you are dealing with. If it's a product or composite function, recognize which rule to apply, like the product rule for products of functions.
Next, differentiate each part separately. For our exercise, differentiate \( u = N \) to get \( u' = 1 \), and differentiate \( v = 1 - \frac{N}{K} \) to get \( v' = -\frac{1}{K} \). Each step of differentiation can often involve basic rules like power or constant rules.
Then, substitute these derivatives back into the rule applied in the first step. Following our steps, this substitution looked like: \[\frac{d}{dN}(N(1-\frac{N}{K})) = 1\cdot(1-\frac{N}{K}) + N\cdot(-\frac{1}{K}).\] By performing these steps systematically, you simplify finding the derivative, even when the function seems complicated at first glance.

Simplification in Calculus

Once you've applied rules like the product rule, simplifying the resulting expression is crucial. Simplification helps in understanding the result better and often makes further calculations easier.
In our derivative example, you’re left with:\[1 - \frac{N}{K} - \frac{N}{K}.\]With simplification, combine like terms to produce a single, easier to interpret expression:

• The two fractions \( -\frac{N}{K} \) combine to make \( -\frac{2N}{K} \).
• Thus, the complete simplified result is \( 1 - \frac{2N}{K} \).

Simplification in calculus is about recognizing patterns and terms that can merge or cancel, reducing complexities in the final expression. It doesn’t change the derivative but clarifies it, showing precisely how the changes in the function behave.