/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Differentiate $$ g(N)=N\left... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

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

Step by step solution

01

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 \( u' = 1 \).
02

Differentiate the Second Function

Now, differentiate the second function \( v = 1 - \frac{N}{K} \). This gives \( v' = -\frac{1}{K} \).
03

Apply Product Rule Components

Substitute into the product rule: \( \frac{d}{dN}(N(1-\frac{N}{K})) = 1\cdot(1-\frac{N}{K}) + N\cdot(-\frac{1}{K}) \).
04

Simplify the Expression

Simplify the expression from step 3: \( 1 - \frac{N}{K} - \frac{N}{K} = 1 - 2\frac{N}{K} \).
05

Final Result

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Assume that a species lives in a habitat that consists of many islands close to a mainland. The species occupies both the mainland and the islands, but, although it is present on the mainland at all times, it frequently goes extinct on the islands. Islands can be recolonized by migrants from the mainland. The following model keeps track of the fraction of islands occupied: Denote the fraction of islands occupied at time \(t\) by \(p(t)\). Assume that each island experiences a constant risk of extinction and that vacant islands (the fraction \(1-p\) ) are colonized from the mainland at a constant rate. Then $$ \frac{d p}{d t}=c(1-p)-e p $$ where \(c\) and \(e\) are positive constants. (a) The gain from colonization is \(f(p)=c(1-p)\) and the loss from extinction is \(g(p)=e p .\) Graph \(f(p)\) and \(g(p)\) for \(0 \leq p \leq 1\) in the same coordinate system. Explain why the two graphs intersect whenever \(e\) and \(c\) are both positive. Compute the point of intersection and interpret its biological meaning. (b) The parameter \(c\) measures how quickly a vacant island becomes colonized from the mainland. The closer the islands, the larger is the value of \(c\). Use your graph in (a) to explain what happens to the point of intersection of the two lines as \(c\) increases. Interpret your result in biological terms.

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