Problem 72 - Algebra and Trigonometry

Question

\(71-76\) m simplify the expression. (This type of expression arises in calculus when using the "quotient rule.") $$ \frac{2 x(x+6)^{4}-x^{2}(4)(x+6)^{3}}{(x+6)^{8}} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{x(-2x + 12)}{(x+6)^5} \).

Step 1: Factor Out Common Terms in the Numerator

In the numerator, identify the common terms in both expressions. Both terms share a factor of \(x(x+6)^3\). We factor this out to get: \[x (x+6)^3 [2(x+6) - 4x]\].

Step 2: Simplify the Expression Inside the Brackets

Inside the brackets, simplify the expression \([2(x+6) - 4x]\). Distribute \(2\) to \((x+6)\), giving \(2x + 12\). Then simplify \(2x + 12 - 4x\) to find \(-2x + 12\).

Step 3: Substitute Simplified Expression Back

Substitute the simplified expression \(-2x + 12\) back into the factored form. The numerator is now \[x(x+6)^3(-2x + 12)\].

Step 4: Further Simplify the Numerator

Distribute \(x(x+6)^3\) across \(-2x + 12\). First, calculate \(-2x \times x(x+6)^3 = -2x^2(x+6)^3\). Next, calculate \(12 \times x(x+6)^3 = 12x(x+6)^3\).

Step 5: Combine Like Terms and Cancel with Denominator

Combine the expressions: \[-2x^2(x+6)^3 + 12x(x+6)^3\]. Factor out \(x(x+6)^3\) to obtain \[x(x+6)^3(-2x + 12)\]. This cancels with the \((x+6)^8\) in the denominator since it reduces to \((x+6)^5\).

Step 6: Write Final Simplified Expression

After canceling, we are left with the simplified expression: \[ \frac{x(-2x + 12)}{(x+6)^5} \].

Key Concepts

Quotient Rule

In calculus, the quotient rule is a method used to find the derivative of a function that is the ratio of two differentiable functions. This rule is essential because it provides a systematic way to deal with these expressions, which arise frequently in differentiation problems.
Here's the basic idea:

• If you have a function in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative is given by:

\[(f/g)' = \frac{g(f') - fg'}{g^2}\]

• This means you take the derivative of the top function \( u \) (numerator) and multiply it by the bottom function \( v \) (denominator), then subtract the derivative of the bottom function multiplied by the top function, all over the bottom function squared.
• This rule helps simplify the complexity of dealing with fractions when finding derivatives.

While the original expression given involves simplification using algebraic techniques, it is derived from functions that typically require the quotient rule for differentiation.

Simplifying Expressions

Simplifying expressions is a crucial skill in algebra and calculus. It involves reducing expressions to their simplest form so they become easier to work with in solving equations or making them ready for further operations.
Key processes when simplifying expressions include:

• Combining like terms: This means adding or subtracting terms that have the same variables raised to the same powers.
• Applying distributive property: This involves multiplying through a set of brackets, such as \( a(b + c) = ab + ac \).
• Canceling common factors: This is especially useful when dealing with fractions, helping reduce the numerator and denominator to their simplest form.

In the original exercise, simplification was critical. By factoring out common terms in the numerator and canceling with the denominator, it became possible to reduce a complex algebraic expression into a much simpler form.

Factoring

Factoring is a method used to break down complex expressions into simpler multiplicative components. This process makes it easier to manipulate and solve expressions or equations in algebra and calculus.
Some common factoring methods include:

• Factoring out the greatest common factor: Identify and extract common factors across terms in an expression.
• Difference of squares: Used when you have two terms that are perfect squares separated by a subtraction sign.
• Factoring trinomials: Breaking down a polynomial into two binomials, often using trials or the quadratic formula.

In the given exercise, factoring was extensively used in the numerator. By recognizing common factors like \( x(x+6)^3 \), it allowed further simplification, eventually leading to a form that could be canceled with parts of the denominator.

Algebra

Algebra is the branch of mathematics that uses symbols and letters to represent numbers and quantities in equations and expressions. It is a foundational tool used not only in solving homework problems but also in various real-world applications.
Algebraic operations include:

• Solving equations: Finding the value of unknowns that satisfy given mathematical statements.
• Simplifying expressions: Reducing complex expressions to their simplest forms.
• Performing functions: Mapping inputs to outputs through well-defined rules.

The provided exercise showcases algebraic operations to simplify a calculus expression by factoring, simplifying, and combining like terms. Understanding basic algebra is crucial since it simplifies not just expressions but also aids in tackling more advanced topics, such as calculus.