Problem 34
Question
Find the derivative of each function by using the Quotient Rule. Simplify your answers. $$ f(x)=\frac{x-1}{x+1} $$
Step-by-Step Solution
Verified Answer
The derivative is \( f'(x) = \frac{2}{(x+1)^2} \).
1Step 1: Identify the functions in the quotient
The given function is \( f(x) = \frac{x-1}{x+1} \). In this case, the numerator function \( u(x) \) is \( x - 1 \), and the denominator function \( v(x) \) is \( x + 1 \).
2Step 2: Differentiate the numerator function
Find the derivative of the numerator function \( u(x) = x - 1 \). The derivative \( u'(x) \) is \( 1 \) because the derivative of \( x \) is \( 1 \) and the derivative of a constant \(-1\) is \( 0 \).
3Step 3: Differentiate the denominator function
Find the derivative of the denominator function \( v(x) = x + 1 \). The derivative \( v'(x) \) is also \( 1 \) for similar reasons: the derivative of \( x \) is \( 1 \), and the derivative of the constant \( 1 \) is \( 0 \).
4Step 4: Apply Quotient Rule
The Quotient Rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). Substitute in the derivatives and functions:\[ f'(x) = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} \]
5Step 5: Simplify the expression
Expand each term in the numerator. First, \( (1)(x+1) = x + 1 \). Then \( (x-1)(1) = x - 1 \). Substitute these into the formula:\[ f'(x) = \frac{x + 1 - x + 1}{(x+1)^2} \]. Simplifying the numerator, we have \( x + 1 - x + 1 = 2 \). So,\[ f'(x) = \frac{2}{(x+1)^2} \].
6Step 6: Present the simplified derivative
The derivative of the function \( f(x) = \frac{x-1}{x+1} \) after simplification is:\[ f'(x) = \frac{2}{(x+1)^2} \].
Key Concepts
Understanding DerivativesSimplification Made EasyWorking with Numerator and Denominator Functions
Understanding Derivatives
Derivatives are a core concept in calculus, used to determine the rate at which a function is changing at any given point.
They show how much a function's output changes with a small change in input, essentially measuring the slope of a function's curve.To find the derivative, we apply rules based on the type of function we have. For example, when working with simple linear functions, the derivative can be quickly calculated by considering that the derivative of a constant is zero and the derivative of a term like \(x\) is 1.
In context, when given \(u(x) = x - 1\), the derivative \(u'(x) = 1\) because the derivative of \(x\) is 1, and any constant’s derivative becomes 0.In the exercise, we needed to find the derivative of a quotient, which requires a specific strategy: the Quotient Rule. Understanding how to apply this rule is key to accurately finding derivatives for such functions.
They show how much a function's output changes with a small change in input, essentially measuring the slope of a function's curve.To find the derivative, we apply rules based on the type of function we have. For example, when working with simple linear functions, the derivative can be quickly calculated by considering that the derivative of a constant is zero and the derivative of a term like \(x\) is 1.
In context, when given \(u(x) = x - 1\), the derivative \(u'(x) = 1\) because the derivative of \(x\) is 1, and any constant’s derivative becomes 0.In the exercise, we needed to find the derivative of a quotient, which requires a specific strategy: the Quotient Rule. Understanding how to apply this rule is key to accurately finding derivatives for such functions.
Simplification Made Easy
Simplification is all about making expressions easier to interpret and work with, without changing their value.
In mathematics, particularly when dealing with derivatives, simplification helps in understanding and presenting the final result clearly.After applying the Quotient Rule, we often end up with a complex expression. The next step is to simplify this expression by combining like terms or using arithmetic operations to reduce it.
In the given exercise, after using the Quotient Rule, we arrived at \(\frac{x + 1 - x + 1}{(x+1)^2}\). By combining terms in the numerator—\(x + 1 - x + 1 = 2\)—we achieve a much simpler form.
This simplification lets us express the derivative as \(\frac{2}{(x+1)^2}\), which is not only easier to understand but also more practical for further calculations or interpretations.
In mathematics, particularly when dealing with derivatives, simplification helps in understanding and presenting the final result clearly.After applying the Quotient Rule, we often end up with a complex expression. The next step is to simplify this expression by combining like terms or using arithmetic operations to reduce it.
In the given exercise, after using the Quotient Rule, we arrived at \(\frac{x + 1 - x + 1}{(x+1)^2}\). By combining terms in the numerator—\(x + 1 - x + 1 = 2\)—we achieve a much simpler form.
This simplification lets us express the derivative as \(\frac{2}{(x+1)^2}\), which is not only easier to understand but also more practical for further calculations or interpretations.
Working with Numerator and Denominator Functions
When applying the Quotient Rule, understanding the roles of the numerator and denominator functions is crucial.
In our context, these are represented by
The rule formula is: if \(f(x) = \frac{u(x)}{v(x)}\), then the derivative \(f'(x)\) is given by:\[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]This formula takes advantage of both the derivatives of the numerator and the denominator. It's carefully set up to ensure that all parts of the function’s behavior are accounted for.
By breaking down the problem and focusing on these two key components—it's easier to apply the rule correctly and simplify the process. Recognizing and correctly calculating the derivatives of both the numerator and denominator, as was done with \(u'(x) = 1\) and \(v'(x) = 1\), simplifies the whole procedure.
In our context, these are represented by
- \(u(x) = x - 1\) as the numerator function
- \(v(x) = x + 1\) as the denominator function
The rule formula is: if \(f(x) = \frac{u(x)}{v(x)}\), then the derivative \(f'(x)\) is given by:\[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]This formula takes advantage of both the derivatives of the numerator and the denominator. It's carefully set up to ensure that all parts of the function’s behavior are accounted for.
By breaking down the problem and focusing on these two key components—it's easier to apply the rule correctly and simplify the process. Recognizing and correctly calculating the derivatives of both the numerator and denominator, as was done with \(u'(x) = 1\) and \(v'(x) = 1\), simplifies the whole procedure.
Other exercises in this chapter
Problem 34
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