Answer:
[tex]\displaystyle y' = -(2x - 5)(6x - 25)[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Distributive Property
Algebra I
- Terms/Coefficients
- Factoring
Calculus
Derivatives
Derivative Notation
Derivative of a constant is 0
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]: [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]
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
y = (2x - 5)²(5 - x)
Step 2: Differentiate
- Derivative Rule [Product Rule]: [tex]\displaystyle y' = \frac{d}{dx}[(2x - 5)^2](5 - x) + (2x - 5)^2\frac{d}{dx}[(5 - x)][/tex]
- Chain Rule [Basic Power Rule]: [tex]\displaystyle y' = [2(2x - 5)^{2 - 1} \cdot \frac{d}{dx}[2x]](5 - x) + (2x - 5)^2\frac{d}{dx}[(5 - x)][/tex]
- Simplify: [tex]\displaystyle y' = [2(2x - 5) \cdot \frac{d}{dx}[2x]](5 - x) + (2x - 5)^2\frac{d}{dx}[(5 - x)][/tex]
- Basic Power Rule: [tex]\displaystyle y' = [2(2x - 5) \cdot 1 \cdot 2x^{1 - 1}](5 - x) + (2x - 5)^2(1 \cdot -x^{1 - 1})][/tex]
- Simplify: [tex]\displaystyle y' = [2(2x - 5) \cdot 2](5 - x) + (2x - 5)^2(-1)[/tex]
- Multiply: [tex]\displaystyle y' = 4(2x - 5)(5 - x) - (2x - 5)^2[/tex]
- Factor: [tex]\displaystyle y' = (2x - 5)[4(5 - x) - (2x - 5)][/tex]
- [Distributive Property] Distribute 4: [tex]\displaystyle y' = (2x - 5)[20 - 4x - (2x - 5)][/tex]
- [Distributive Property] Distribute negative: [tex]\displaystyle y' = (2x - 5)[20 - 4x - 2x + 5][/tex]
- [Subtraction] Combine like terms (x): [tex]\displaystyle y' = (2x - 5)[20 - 6x + 5][/tex]
- [Addition] Combine like terms: [tex]\displaystyle y' = (2x - 5)(25 - 6x)[/tex]
- Factor: [tex]\displaystyle y' = -(2x - 5)(6x - 25)[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e
