Problem 73 - Algebra and Trigonometry

Question

\(71-76\) m simplify the expression. (This type of expression arises in calculus when using the "quotient rule.") $$ \frac{2(1+x)^{1 / 2}-x(1+x)^{-1 / 2}}{x+1} $$

Step-by-Step Solution

Answer

The simplified expression is \((1+x)^{-3/2}(2 + x)\).

Step 1: Identify the Expression

We start by focusing on the given expression \( \frac{2(1+x)^{1/2} - x(1+x)^{-1/2}}{x+1} \). Our goal is to simplify it.

Step 2: Simplify the Numerator

The numerator is \( 2(1+x)^{1/2} - x(1+x)^{-1/2} \). We should try to factor out a common factor if possible. Notice that \((1+x)^{-1/2}\) is common: \[((1+x)^{-1/2})(2(1+x) - x)\] Further simplification of the expression inside the parentheses results in \(2 + 2x - x = 2 + x\), so: \[2(1+x)^{1/2} - x(1+x)^{-1/2} = ((1+x)^{-1/2})(2 + x)\]

Step 3: Form the Quotient

Now substitute back into the original expression:\[\frac{((1+x)^{-1/2})(2 + x)}{x+1}\]

Step 4: Cancel Common Terms

Since \(x+1\) and \(1+x\) are the same, \(x+1\) cancels with part of \((1+x)\) in the numerator:\[\frac{(1+x)^{-1/2}(2 + x)}{x+1} = (1+x)^{-3/2}(2 + x)\]

Step 5: Final Simplified Expression

Thus, the simplified expression is:\[(1+x)^{-3/2}(2 + x)\]

Key Concepts

Quotient Rule

The Quotient Rule is a technique used in calculus to find the derivative of a quotient of two functions. If you have two differentiable functions, say \( u(x) \) and \( v(x) \), the derivative of their quotient is given by:

• \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

The formula helps in calculating the rate of change of the quotient of two functions.
Considering the original exercise where we have a complex fraction, applying the Quotient Rule might initially look helpful; however, the problem focuses on algebraic simplification rather than differentiation.
Simplifying algebraic expressions like the one in the original exercise prepares students for using the Quotient Rule by making the expressions easier to work with, especially those seen in calculus.

Exponents

Exponents are a mathematical notation indicating the number of times a number is multiplied by itself. They are essential in simplifying expressions involving powers, and it's important to understand how they work.
In this exercise, you see exponents such as \((1+x)^{1/2}\) and \((1+x)^{-1/2}\). These are fractional exponents, where the exponent \(1/2\) represents a square root, and \(-1/2\) means reciprocal squared root.

• \((1+x)^{1/2} = \sqrt{1+x}\)
• \((1+x)^{-1/2} = \frac{1}{\sqrt{1+x}}\)

Understanding how to manipulate these helps simplify expressions and is critical for more advanced calculus operations.
Recognizing when terms can be factored out using properties of exponents reduces complexity, as shown in the simplification where \((1+x)^{-1/2}\) is factored.

Factoring

Factoring is a method of breaking down algebraic expressions into simpler 'factors' that can be easily manipulated and understood.
When simplifying an expression like the one in the exercise, finding common factors in the terms helps to simplify operations.
In our solution, we factored out \((1+x)^{-1/2}\) from the numerator, which clarified the expression and made it easier to proceed with simplifying the fraction.

• Factoring turns expressions like \(2(1+x)^{1/2} - x(1+x)^{-1/2}\) into \((1+x)^{-1/2}(2 + x)\).

This process is an essential skill for students to master before tackling more complicated algebraic manipulations and calculus problems.
Effective factoring significantly reduces the computation needed and helps in understanding the underlying structure of the expression.

Calculus Applications

In calculus, simplifying algebraic expressions serves as a foundational skill for more complex problem-solving scenarios.
Expressions of the form in our exercise commonly arise in differentiation, especially with the Quotient Rule, and integration processes where simplification can make calculations more manageable.
The simplified expression \((1+x)^{-3/2}(2 + x)\) reveals operations like differentiation or integration might be required for calculus applications, especially when calculating areas or solving differential equations.

• The structure \((1+x)^{-3/2}\) might hint at integration using substitution or partial fractions.
• Simplification aids in cleanly applying rules like the Chain Rule alongside the Quotient Rule.

Algebraic simplification helps students build the intuition necessary to deal with real-world problems efficiently and provides the groundwork for deeper calculus learning.