Answer:
b. [tex]\displaystyle \frac{1}{2}[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Algebra I
- Functions
- Function Notation
- Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
- Exponential Rule [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]
Calculus
Derivatives
Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle H(x) = \sqrt[3]{F(x)}[/tex]
Step 2: Differentiate
- Rewrite function [Exponential Rule - Root Rewrite]: [tex]\displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}[/tex]
- Chain Rule: [tex]\displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)][/tex]
- Basic Power Rule: [tex]\displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)[/tex]
- Simplify: [tex]\displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}[/tex]
- Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}[/tex]
Step 3: Evaluate
- Substitute in x [Derivative]: [tex]\displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}[/tex]
- Substitute in function values: [tex]\displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}[/tex]
- Exponents: [tex]\displaystyle H'(5) = \frac{6}{3(4)}[/tex]
- Multiply: [tex]\displaystyle H'(5) = \frac{6}{12}[/tex]
- Simplify: [tex]\displaystyle H'(5) = \frac{1}{2}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e
