Problem 37
Question
Differentiate $$ g(N)=N\left(1-\frac{N}{K}\right) $$ with respect to \(N\). Assume that \(K\) is a positive constant.
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{dg}{dN} = 1 - \frac{2N}{K} \).
1Step 1: Write Down the Function
The given function is \( g(N) = N\left(1 - \frac{N}{K}\right) \). We need to differentiate this function with respect to \( N \).
2Step 2: Distribute the Terms
Simplify the function by distributing the \( N \) through the parentheses: \( g(N) = N - \frac{N^2}{K} \).
3Step 3: Differentiate Each Term Separately
Differentiate each term with respect to \( N \). For the first term, \( \frac{d}{dN}[N] = 1 \). For the second term, \( \frac{d}{dN}\left[-\frac{N^2}{K}\right] = -\frac{d}{dN}\left[\frac{N^2}{K}\right] = -\frac{2N}{K} \) since \( K \) is a constant.
4Step 4: Combine the Derivatives
Combine the results from differentiating each term: the derivative \( \frac{dg}{dN} = 1 - \frac{2N}{K} \).
Key Concepts
Understanding ConstantsDerivative BasicsSimplifying Expressions
Understanding Constants
In mathematics, a constant is a value that does not change. When dealing with differentiation, recognizing constants is crucial because they simplify the process.
In the expression given in the original exercise, the term 'K' is described as a constant. Unlike variables that can change, constants like 'K' are fixed values.
When differentiating expressions that include constants, remember that the constant itself doesn't vary with the variable you are differentiating with respect to. Thus, in expressions like \( \frac{N^2}{K} \), 'K' remains unaffected when finding the derivative.
In the expression given in the original exercise, the term 'K' is described as a constant. Unlike variables that can change, constants like 'K' are fixed values.
When differentiating expressions that include constants, remember that the constant itself doesn't vary with the variable you are differentiating with respect to. Thus, in expressions like \( \frac{N^2}{K} \), 'K' remains unaffected when finding the derivative.
- Example: Differentiating \( c \cdot x \) where 'c' is a constant, gives the derivative \( c \).
- Constants simplify differentiation as they can be factored out or ignored according to the rules of derivatives.
Derivative Basics
A derivative represents how a function changes as its input changes. It is the backbone of calculus, especially when considering rates of change.
In the provided exercise, we aim to differentiate the function \( g(N) = N \left( 1 - \frac{N}{K} \right) \) with respect to \( N \).
Differentiation involves computing the rate at which the function \( g(N) \) changes as \( N \) changes.
In the provided exercise, we aim to differentiate the function \( g(N) = N \left( 1 - \frac{N}{K} \right) \) with respect to \( N \).
Differentiation involves computing the rate at which the function \( g(N) \) changes as \( N \) changes.
- The basic rule is: the derivative of \( x^n \) is \( nx^{n-1} \).
- For a constant \( c \), the derivative is 0 because constants never change.
- Differentiate \( N \) to obtain 1, as \( N^1 \) becomes \( 1 \cdot N^{0} = 1 \).
- For \( -\frac{N^2}{K} \), treat \( K \) as a constant, giving \(-\frac{2N}{K}\).
Simplifying Expressions
Simplification is an important step in making derivative calculations more straightforward. It involves rewriting functions in a simpler form before any math operations, like differentiation, are applied.
The original function \( g(N) = N \left( 1 - \frac{N}{K} \right) \) can be difficult to handle directly.
By distributing \( N \) through the parentheses, the expression becomes \( g(N) = N - \frac{N^2}{K} \).
The original function \( g(N) = N \left( 1 - \frac{N}{K} \right) \) can be difficult to handle directly.
By distributing \( N \) through the parentheses, the expression becomes \( g(N) = N - \frac{N^2}{K} \).
- Distributing a term means you multiply it with everything inside the parentheses.
- This reduces complex expressions into simpler, individual terms.
- Clarifying the terms to be differentiated.
- Making calculations easier and less error-prone.
- Allowing for straightforward application of differentiation rules.
Other exercises in this chapter
Problem 37
Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \frac{x}{x+1} $$
View solution Problem 37
Assume that \(f(x)\) is differentiable. Find an expression for the derivative of \(y\) at \(x=1\), assuming that \(f(1)=2\) and \(f^{\prime}(1)=-1\) \(y=2 x f(x
View solution Problem 37
In Problems 36-39, assume that \(f(x)\) and \(g(x)\) are differentiable. Find \(\frac{d}{d x}\left(\frac{f(x)}{g(x+1)}\right)\)
View solution Problem 38
A measurement error in \(x\) affects the accuracy of the value \(f(x) .\) In each case, determine an interval of the form $$[f(x)-\Delta f, f(x)+\Delta f]$$ tha
View solution