Chapter 2: Problem 14
$$\text { Find } f^{\prime}(x)$$ $$f(x)=\sqrt[3]{\frac{8}{x}}$$
Short Answer
Expert verified
\( f'(x) = -\frac{8}{3x^{8/3}} \)
Step by step solution
01
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 \( f(x) = \left(\frac{8}{x}\right)^{1/3} \).
02
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 \( \frac{1}{3} \): \( f(x) = (8x^{-1})^{1/3} \).
03
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 \( \frac{d}{dx} (u^{1/3}) = \frac{1}{3}u^{-2/3} \cdot \frac{du}{dx} \).
04
Differentiate the Inner Function
Find \( \frac{du}{dx} \) for \( u = 8x^{-1} \). Use the power rule: \( \frac{d}{dx}(8x^{-1}) = 8 \cdot (-1)x^{-2} = -8x^{-2} \).
05
Combine Derivatives
Substitute back into the chain rule result: \( f'(x) = \frac{1}{3}(8x^{-1})^{-2/3} \cdot (-8x^{-2}) \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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: \( \frac{d}{dx}(8x^{-1}) = -8x^{-2} \)
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 \( x^{-1} \).
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.