Problem 14 - Calculus - AP Edition

Question

$$\text { Find } f^{\prime}(x)$$ $$f(x)=\sqrt[3]{\frac{8}{x}}$$

Step-by-Step Solution

Answer

\( f'(x) = -\frac{8}{3x^{8/3}} \)

Step 1: Simplify the Function

First, rewrite the function with the exponent notation. The cube root can be expressed as a fractional exponent. Therefore, rewrite the function \( f(x) = \sqrt[3]{\frac{8}{x}} \) as \( f(x) = \left(\frac{8}{x}\right)^{1/3} \).

Step 2: Apply the Quotient Exponent Rule

Use the properties of exponents to simplify. The expression \( \frac{8}{x} \) can be written as \( 8x^{-1} \). Thus, raise this to the power of \( \frac{1}{3} \): \( f(x) = (8x^{-1})^{1/3} \).

Step 3: Apply the Chain Rule

For the derivative of \( (8x^{-1})^{1/3} \), use the chain rule. Let \( u = 8x^{-1} \), so that \( f(x) = u^{1/3} \). Then, the derivative is \( \frac{d}{dx} (u^{1/3}) = \frac{1}{3}u^{-2/3} \cdot \frac{du}{dx} \).

Step 4: Differentiate the Inner Function

Find \( \frac{du}{dx} \) for \( u = 8x^{-1} \). Use the power rule: \( \frac{d}{dx}(8x^{-1}) = 8 \cdot (-1)x^{-2} = -8x^{-2} \).

Step 5: Combine Derivatives

Substitute back into the chain rule result: \( f'(x) = \frac{1}{3}(8x^{-1})^{-2/3} \cdot (-8x^{-2}) \).

Step 6: Simplify

Simplify the expression: \( f'(x) = \frac{-8}{3}(8x^{-1})^{-2/3} \cdot x^{-2} \). Recall \( (8x^{-1})^{-2/3} = (\frac{x}{8})^{2/3} \), and simplify to find \( f'(x) = \frac{-8}{3} \cdot (\frac{x}{8})^{2/3} \cdot x^{-2} \). Combine the terms to get the final derivative.

Key Concepts

Chain Rule

The chain rule is a technique used in calculus to differentiate composite functions easily. When dealing with a function that is nested within another function, the chain rule becomes essential. In this exercise, we are given a function in the form of \( f(x) = (8x^{-1})^{1/3} \), where the inner function is \( u = 8x^{-1} \). The outer function is \( u^{1/3} \). To find the derivative \( f'(x) \), the chain rule guides us to first differentiate the outer function with respect to the inner function \( u \), and then multiply by the derivative of the inner function with respect to \( x \). Here is how it would look:

• Differentiate the outer function: \( \frac{d}{du}(u^{1/3}) = \frac{1}{3}u^{-2/3} \)
• Differentiate the inner function: \( \frac{d}{dx}(8x^{-1}) = -8x^{-2} \)

To complete the chain rule: \( f'(x) = \frac{1}{3}u^{-2/3} \cdot (-8x^{-2}) \). This method turns a potentially complex differentiation problem into a manageable process, emphasizing the utility of the chain rule in calculus.

Exponent Rules

Exponent rules are crucial in simplifying expressions and are especially helpful in differentiation problems such as this one. When dealing with the function \( f(x) = \sqrt[3]{\frac{8}{x}} \), using exponent rules makes the task easier. A radical expression can be rewritten using fractional exponents, allowing the application of different rules for differentiation:

• Converting radical to fractional exponent: \( \sqrt[3]{\frac{8}{x}} = (\frac{8}{x})^{1/3} \).
• Using the quotient rule for exponents: \( (\frac{8}{x}) = 8x^{-1} \), which utilizes the principle that division by \( x \) is equivalent to multiplying by \( x^{-1} \).

Understanding these rules is vital for tackling more complex algebraic manipulations. It helps transform expressions into forms amenable to differentiation, leading to more efficient and accurate solutions.

Power Rule

The power rule is a fundamental tool in differentiation, simplifying the process of finding derivatives of functions that are powers of \( x \). Applied to functions like \( 8x^{-1} \), the power rule states: \( \frac{d}{dx}(x^n) = nx^{n-1} \). For the inner function \( u = 8x^{-1} \), we apply the power rule:

• The coefficient \( 8 \) is preserved during differentiation.
• For \( x^{-1} \), the derivative becomes \( -x^{-2} \) or \( -8 \cdot x^{-2} \) when adjusted with the coefficient.

Applying the power rule reduces the complexity of finding the derivative of polynomial and rational expressions, emphasizing its role in efficient differentiation, as seen in the step-by-step solution. By ensuring mastery of the power rule, students can tackle a wide range of differential calculus problems.