Problem 33 - Brief Applied Calculus

Question

31-38. Find the indicated derivatives. If \(f(x)=6 \sqrt[3]{x^{2}}-\frac{48}{\sqrt[3]{x}}\), find \(f^{\prime}(8)\).

Step-by-Step Solution

Answer

The value of \( f'(8) \) is 3.

Step 1: Rewrite the function

The original function is given as \( f(x) = 6 \sqrt[3]{x^2} - \frac{48}{\sqrt[3]{x}} \). To differentiate, rewrite the expression in terms of exponents. \( \sqrt[3]{x^2} \) can be rewritten as \( x^{2/3} \), and \( \frac{1}{\sqrt[3]{x}} \) as \( x^{-1/3} \). Thus, \( f(x) = 6x^{2/3} - 48x^{-1/3} \).

Step 2: Differentiate the function

To find \( f'(x) \), apply the power rule for each term. The derivative of \( 6x^{2/3} \) is \( (6) \cdot \frac{2}{3} \cdot x^{-1/3} = 4x^{-1/3} \). The derivative of \( -48x^{-1/3} \) is \( (-48) \cdot (-\frac{1}{3}) \cdot x^{-4/3} = 16x^{-4/3} \). So, \( f'(x) = 4x^{-1/3} + 16x^{-4/3} \).

Step 3: Simplify the derivative

The expression for \( f'(x) \) is \( 4x^{-1/3} + 16x^{-4/3} \). This can be simplified for evaluation: \( f'(x) = \frac{4}{x^{1/3}} + \frac{16}{x^{4/3}} \).

Step 4: Evaluate the derivative at x = 8

Substitute \( x = 8 \) into the simplified derivative. Calculate \( f'(8) = \frac{4}{8^{1/3}} + \frac{16}{8^{4/3}} \). Since \( 8^{1/3} = 2 \), \( f'(8) = \frac{4}{2} + \frac{16}{16} = 2 + 1 = 3 \).

Key Concepts

Power Rule

The power rule is a fundamental technique in calculus used to find the derivative of functions involving exponents. It states that if you have a function of the form \( f(x) = ax^n \), its derivative will be \( f'(x) = anx^{n-1} \). This means you multiply the coefficient \( a \) by the exponent \( n \), and then subtract one from the exponent. For example, take the function \( f(x) = 6x^{2/3} \). Using the power rule, you multiply 6 by \( \frac{2}{3} \) and subtract 1 from the exponent, resulting in \( 4x^{-1/3} \). Similarly, for \( -48x^{-1/3} \), you multiply -48 by \( \frac{-1}{3} \) and decrease the exponent by 1, giving \( 16x^{-4/3} \).

• This rule simplifies the process of differentiation for polynomial-like functions.
• It allows for quick calculation of derivatives, which are fundamental in understanding rates of change and slopes of functions.

Utilizing the power rule is essential for efficiently solving differentiation problems, particularly those involving roots and fractions expressed as powers.

Differentiation

Differentiation is the mathematical process of finding a derivative, which measures how a function changes as its input changes. In simpler terms, the derivative tells us the rate at which one quantity changes with respect to another. It is a core concept in calculus used to understand motion, growth, and other dynamic systems.In the given problem, differentiation is applied to the function \( f(x) = 6 \sqrt[3]{x^2} - \frac{48}{\sqrt[3]{x}} \). The process starts by converting radicals to fractional exponents, allowing for the use of the power rule.

• Understanding differentiation involves recognizing how to rewrite expressions for easier derivative calculation.
• It is used across various fields such as physics, engineering, and economics for analyzing changing conditions.

Differentiation transforms complex expressions into simpler forms conducive to analyzing and solving real-world problems.

Calculus Problem Solving

Calculus problem solving involves applying mathematical techniques to find solutions in dynamic situations. It requires breaking down problems into manageable steps, often starting with rewriting expressions for better clarity. In this exercise, the goal is to find the derivative of a complex function and evaluate it at a specific point.Follow these steps:- **Rewrite the function**: Use algebraic manipulations to express the function clearly, as seen by converting roots to power forms.- **Apply differentiation techniques**: Use rules like the power rule to find derivatives of individual terms.- **Simplify and evaluate**: Simplify the derivative expression for easier mathematical operations and evaluate at the given point (e.g., using \( x = 8 \)).These steps illustrate:

• How algebra and calculus tools work in unison to solve intricate problems.
• The importance of methodical approaches in reaching precise solutions quickly.

Mastery in calculus problem solving enhances mathematical abilities and equips learners with strategies for tackling diverse career-related challenges.