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Find the derivatives of the functions. \begin{equation} y=\frac{(x+1)(x+2)}{(x-1)(x-2)} \end{equation}

Short Answer

Expert verified
The derivative is \( y' = \frac{-6x^2 + 12x + 15}{(x^2 - 3x + 2)^2} \).

Step by step solution

01

Identify the Function Type

The given function is a rational function, specifically of the form \( \frac{u(x)}{v(x)} \), where \( u(x) = (x+1)(x+2) \) and \( v(x) = (x-1)(x-2) \).
02

Differentiate Using the Quotient Rule

Use the quotient rule for differentiation, which states that if \( y = \frac{u(x)}{v(x)} \), then \( y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). First, we need to find \( u'(x) \) and \( v'(x) \).
03

Find the Derivative of the Numerator

The numerator \( u(x) = (x+1)(x+2) = x^2 + 3x + 2 \). By differentiating term-by-term, we find \( u'(x) = 2x + 3 \).
04

Find the Derivative of the Denominator

The denominator \( v(x) = (x-1)(x-2) = x^2 - 3x + 2 \). Differentiating term-by-term gives \( v'(x) = 2x - 3 \).
05

Apply the Quotient Rule

Plug \( u(x), u'(x), v(x), \) and \( v'(x) \) into the quotient rule: \[y' = \frac{(2x + 3)(x^2 - 3x + 2) - (x^2 + 3x + 2)(2x - 3)}{(x^2 - 3x + 2)^2}\]
06

Simplify the Expression

Expand and simplify the numerator: \[y' = \frac{[2x^3 - 6x^2 + 4x + 3x^2 - 9x + 6] - [2x^3 - 3x^2 + 4x - 6x - 9]}{(x^2 - 3x + 2)^2}\]Combine like terms:\[y' = \frac{-6x^2 + 12x + 15}{(x^2 - 3x + 2)^2}\]
07

Finalize the Simplified Derivative

Double-check each term of the expanded form to ensure correctness and consistency. The final simplified derivative is:\[y' = \frac{-6x^2 + 12x + 15}{(x^2 - 3x + 2)^2}\]

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Key Concepts

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

Rational Functions
Rational functions are a fundamental concept in algebra and calculus, which involve the division of two polynomials. In simplest terms, if you have a function expressed as \( y = \frac{u(x)}{v(x)} \), where \( u(x) \) and \( v(x) \) are both polynomials, you have a rational function. Understanding this type of function is essential because it enables a wide range of mathematical analysis including finding limits, asymptotes, and derivatives.

These functions often exhibit interesting behavior, such as vertical and horizontal asymptotes, which reflect their essential nature.
  • Vertical asymptotes occur where the denominator is zero.
  • Horizontal asymptotes can help understand the end behavior as \( x \to \infty \) or \( x \to -\infty \).
Analyzing rational functions gives insights into how changes in the numerator and denominator affect overall behavior.
Quotient Rule
The quotient rule is a guideline in calculus used to differentiate functions that are the quotient of two other functions. When you have a function \( y = \frac{u(x)}{v(x)} \), the derivative \( y' \) can be found using the quotient rule:\[ y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]This rule is quite powerful because it gives a structured approach to tackle the differentiation of rational functions. The resources required include:
  • Knowing both the numerator and the denominator functions, \( u(x) \) and \( v(x) \).
  • Finding their respective derivatives, \( u'(x) \) and \( v'(x) \).
Thus, always begin by identifying components and their derivatives, then correctly apply them in the formula. This ensures an accurate calculation of the derivative.
Differentiation
Differentiation is the process of finding a derivative, which indicates the rate of change of a function concerning one of its variables. It's a cornerstone concept in calculus principles.

Derivatives provide critical information including velocity in physics when the function in question represents displacement. When you're dealing with more complex functions, like rational functions, learning to differentiate each term correctly is crucial.
  • Differentiation can simplify expressions, which assists in finding rates of change easily in various real-world applications.
  • Applying different rules of differentiation helps to systematically deal with multiple forms, from simple polynomials to complex quotients.
Grasping the techniques of differentiation empowers you to explore new equations and functions with confidence.
Calculus
Calculus, a branch of mathematics, exists to study continuous change. It's composed of two main ideas: differentiation and integration. Differentiation, as discussed, provides rates of change, while integration allows for the calculation of areas under curves and accumulation functions.

Calculus bridges the gap between algebraic concepts and practical problems. Its methods are essential in various fields like engineering, physics, economics, and more.
  • Understanding calculus enables you to solve for everything from optimizing industrial processes to calculating celestial bodies' orbits.
  • Learning calculus concepts offers tools that significantly enhance problem-solving skills and analytical thinking.
Finally, mastering calculus fundamentals is invaluable for any scientific or technical endeavors.
Numerator and Denominator Derivatives
In handling rational functions, it is vital to understand both the numerator derivative \( u'(x) \) and the denominator derivative \( v'(x) \).

The process of finding these derivatives involves:
  • Taking each polynomial separately and differentiating term by term.
  • Changing only one component at a time simplifies the calculation process and minimizes errors.
For example, if you have the numerator \( u(x) = (x+1)(x+2) \) expanded to \( u(x) = x^2 + 3x + 2 \), its derivative is \( u'(x) = 2x + 3 \). Similarly, for the denominator \( v(x) = (x-1)(x-2) \) expanded to \( v(x) = x^2 - 3x + 2 \), its derivative is \( v'(x) = 2x - 3 \). These calculated derivatives are crucial to correctly apply the quotient rule for differentiation. By focusing on each polynomial separately, you ensure a structured approach.

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Most popular questions from this chapter

In Exercises \(41-58\) find \(d y / d t\) $$ y=(1+\cos 2 t)^{-4} $$

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A growing raindrop Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate.

Cardiac output In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 \(\mathrm{L} / \mathrm{min.}\) . At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min}\) . If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min.}\) Your cardiac output can be calculated with the formula $$y=\frac{Q}{D}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{ml} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{ml} / \mathrm{min}\) and \(D=97-56=41 \mathrm{ml} / \mathrm{L}\) $$y=\frac{233 \mathrm{ml} / \mathrm{min}}{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

If \(r=\sin (f(t)), f(0)=\pi / 3,\) and \(f^{\prime}(0)=4,\) then what is \(d r / d t\) at \(t=0 ?\)

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