Answer:
[tex]\displaystyle y'(1, \frac{3}{2}) = -3[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Algebra I
- Coordinates (x, y)
- Functions
- Function Notation
- Terms/Coefficients
- Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
Calculus
Derivatives
Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Step-by-step explanation:
Step 1: Define
[tex]\displaystyle y = \frac{3}{2x^2}[/tex]
[tex]\displaystyle \text{Point} \ (1, \frac{3}{2})[/tex]
Step 2: Differentiate
- [Function] Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle y = \frac{3}{2}x^{-2}[/tex]
- Basic Power Rule: [tex]\displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}[/tex]
- Simplify: [tex]\displaystyle y' = -3x^{-3}[/tex]
- Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle y' = \frac{-3}{x^3}[/tex]
Step 3: Solve
- Substitute in coordinate [Derivative]: [tex]\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}[/tex]
- Evaluate exponents: [tex]\displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}[/tex]
- Divide: [tex]\displaystyle y'(1, \frac{3}{2}) = -3[/tex]
Topic: AP Calculus AB/BC (Calculus I/II)
Unit: Derivatives
Book: College Calculus 10e
