Problem 27 - Brief Applied Calculus

Question

Find the derivative of each function in two ways: a. Using the Quotient rule. b. Simplifying the original function and using the Power Rule. Your answers to parts (a) and (b) should agree. $$ \frac{x^{8}}{x^{2}} $$

Step-by-Step Solution

Answer

The derivative is \( 6x^5 \) using both methods.

Step 1: Identify the Quotient Rule Formula

The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = x^8 \) and \( v = x^2 \).

Step 2: Differentiate Using the Quotient Rule

First, find the derivatives \( u' \) and \( v' \). Since \( u = x^8 \), \( u' = 8x^7 \). Since \( v = x^2 \), \( v' = 2x \). Apply the quotient rule: \[ (x^8/x^2)' = \frac{8x^7 \cdot x^2 - x^8 \cdot 2x}{(x^2)^2} \]. Simplifying, \[ = \frac{8x^9 - 2x^9}{x^4} = \frac{6x^9}{x^4} = 6x^5 \].

Step 3: Simplify the Original Function

Before differentiation, simplify \( \frac{x^8}{x^2} \) to \( x^{8-2} = x^6 \).

Step 4: Differentiate the Simplified Function Using the Power Rule

Apply the power rule to the simplified function \( x^6 \) which states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). Therefore, \( \frac{d}{dx}[x^6] = 6x^5 \).

Step 5: Verify Consistency in Answers

Both methods produced the same derivative: \( 6x^5 \). Thus, the computations are consistent and correct.

Key Concepts

Quotient Rule

The quotient rule is a fundamental technique in calculus employed for differentiating functions that are expressed as fractions, specifically when one function is divided by another. Imagine you have a complex function structured as \( \frac{u}{v} \). To find its derivative, you'd use the formula:
\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}.\]This method systematically helps in determining how changes in both the numerator's and the denominator's functions affect the rate of change of the entire function itself. From our example, we identified \( u = x^8 \) and \( v = x^2 \). By differentiating these, we found \( u' = 8x^7 \) and \( v' = 2x \). Plugging these into the quotient rule gives us:

• \( u'v = 8x^7 \cdot x^2 = 8x^9 \)
• \( uv' = x^8 \cdot 2x = 2x^9 \)
• Thus, the derivative becomes \( \frac{8x^9 - 2x^9}{x^4} = \frac{6x^9}{x^4} = 6x^5 \)

This shows the power of the quotient rule, providing a structured approach to such differentiation problems.

Power Rule

The power rule is one of the simplest yet most useful rules for finding derivatives of polynomial expressions. It states that if you have a function \( x^n \), its derivative is calculated as:
\[\frac{d}{dx}[x^n] = nx^{n-1}.\]This rule simplifies the process by bringing down the exponent as a coefficient and reducing the exponent by one. For instance, with the simplified function \( x^6 \) derived from \( \frac{x^8}{x^2} = x^6 \), applying the power rule gives:
\[\frac{d}{dx}[x^6] = 6x^5,\]which matches the result obtained using the quotient rule. This consistency confirms the correctness of both approaches. The power rule is straightforward and is often the quickest method if you can simplify the equation before differentiation.

Simplifying Functions

Simplifying a function before differentiating can transform complex problems into simpler ones, potentially saving time and reducing errors. The approach involves reducing the complexity of the expression, either by canceling common terms or combining like terms. In our problem, the original fraction \( \frac{x^8}{x^2} \) simplifies directly to \( x^{8-2} = x^6 \). This reduction makes it easier to apply the power rule as opposed to dealing with a more complicated fraction.When you simplify first, you often make the derivative process much more intuitive. Furthermore, simplification can help clarify solutions, making them easier to check and understand. This approach is especially powerful when dealing with algebraic expressions where terms can be directly reduced before moving forward to differentiation.