Chapter 23: Problem 34
Find the derivative of each function.. $$y=\frac{x-1}{x+1}$$
Short Answer
Expert verified
\(y' = \frac{2}{(x+1)^2}\)
Step by step solution
01
Identify the function
The function given is a rational function, which is a division of two functions of x: the numerator function, which is x - 1, and the denominator function, which is x + 1.
02
Apply the quotient rule
The quotient rule states that if you have a function that is the division of two other functions, in this case, u = x - 1 and v = x + 1, the derivative of the function is given by: \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\cdot\frac{du}{dx} - u\cdot\frac{dv}{dx}}{v^2}\). Therefore, we need to differentiate both u and v with respect to x.
03
Differentiate the numerator and denominator
The derivative of the numerator (u = x - 1) with respect to x is 1, since the derivative of x is 1 and the derivative of a constant is 0. The derivative of the denominator (v = x + 1) with respect to x is also 1 for the same reasons.
04
Apply the derivatives to the quotient rule
Using the derivatives from Step 3 and applying the quotient rule, we get: \(\frac{d}{dx}\left(\frac{x-1}{x+1}\right) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2}\). This simplifies to \(\frac{x+1 - (x-1)}{(x+1)^2}\).
05
Simplify the derivative expression
Simplify the numerator by distributing the subtraction and combining like terms: \(x + 1 - x + 1 = 2\). The derivative is then \(\frac{2}{(x+1)^2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Function Differentiation
Differentiating rational functions is often encountered in calculus, and it involves taking the derivative of a function that is composed of a ratio of two polynomials. A rational function is typically of the form \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are polynomial functions of \( x \).
To differentiate a rational function, the quotient rule comes into play, which is a method used when dividing two functions. This rule is expressed as \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\cdot\frac{du}{dx} - u\cdot\frac{dv}{dx}}{v^2} \), where \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) represent the derivatives of \( u \) and \( v \) with respect to \( x \), respectively. The use of the quotient rule ensures that we accurately account for the rate of change of both the numerator and the denominator.
For the given exercise, the function \( y = \frac{x-1}{x+1} \) is differentiated using this exact technique, showcasing the practical application of the quotient rule for rational functions.
To differentiate a rational function, the quotient rule comes into play, which is a method used when dividing two functions. This rule is expressed as \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\cdot\frac{du}{dx} - u\cdot\frac{dv}{dx}}{v^2} \), where \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) represent the derivatives of \( u \) and \( v \) with respect to \( x \), respectively. The use of the quotient rule ensures that we accurately account for the rate of change of both the numerator and the denominator.
For the given exercise, the function \( y = \frac{x-1}{x+1} \) is differentiated using this exact technique, showcasing the practical application of the quotient rule for rational functions.
Derivative of a Function
The derivative of a function at a point is the rate at which the value of the function is changing at that point. It is one of the fundamental concepts of calculus and serves as the cornerstone for understanding rates of change and slopes of curves. The process of finding a derivative is called differentiation.
There are various rules to find derivatives, such as the power rule (for polynomials), product rule (for products of functions), chain rule (for composite functions), and as seen in the exercise, the quotient rule for rational functions. The derivative represents the instant slope of the function and is denoted as \( f'(x) \) or \( \frac{df}{dx} \).
In the context of the exercise provided, the differentiation of the simple polynomials \( u = x - 1 \) and \( v = x + 1 \) is straightforward and each yields a derivative of 1, according to the power rule. These derivatives are then applied in the quotient rule to find the derivative of the overall rational function.
There are various rules to find derivatives, such as the power rule (for polynomials), product rule (for products of functions), chain rule (for composite functions), and as seen in the exercise, the quotient rule for rational functions. The derivative represents the instant slope of the function and is denoted as \( f'(x) \) or \( \frac{df}{dx} \).
In the context of the exercise provided, the differentiation of the simple polynomials \( u = x - 1 \) and \( v = x + 1 \) is straightforward and each yields a derivative of 1, according to the power rule. These derivatives are then applied in the quotient rule to find the derivative of the overall rational function.
Simplifying Derivatives
Simplifying derivatives is a critical final step in differentiation that helps in understanding and working with the derived function more conveniently. It often involves algebraic manipulation such as expanding, factoring, canceling terms, and combining like terms.
After applying rules of differentiation, such as the quotient rule, the resulting expression can be large or complex. Simplification reduces such expressions to their most manageable forms. This usually makes subsequent calculus operations like finding stationary points, solving equations, or integrating much clearer.
In the given exercise, simplification is evident in the final step where subtracting and combining like terms in the numerator \( x+1 - (x-1) \) simplifies to 2. Therefore, the derivative \( \frac{2}{(x+1)^2} \) is the simplified form of the function’s derivative, making it easier to understand and use in further problems or applications.
After applying rules of differentiation, such as the quotient rule, the resulting expression can be large or complex. Simplification reduces such expressions to their most manageable forms. This usually makes subsequent calculus operations like finding stationary points, solving equations, or integrating much clearer.
In the given exercise, simplification is evident in the final step where subtracting and combining like terms in the numerator \( x+1 - (x-1) \) simplifies to 2. Therefore, the derivative \( \frac{2}{(x+1)^2} \) is the simplified form of the function’s derivative, making it easier to understand and use in further problems or applications.
