Answer:
[tex]\displaystyle y' = (1 + x)^2(4x + 1)[/tex]
General Formulas and Concepts:
Algebra I
- Terms/Coefficients
- Functions
- Function Notation
- Factoring
Calculus
Derivatives
Derivative Notation
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
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]
Explanation:
Step 1: Define
Identify
y = x(1 + x)³
Step 2: Differentiate
- Product Rule [Derivative Rule - Chain Rule]: [tex]\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot \frac{d}{dx}[1 + x][/tex]
- Derivative Property [Addition/Subtraction]: [tex]\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot (\frac{d}{dx}[1] + \frac{d}{dx}[x])[/tex]
- Basic Power Rule: [tex]\displaystyle y' = x^{1 - 1} \cdot (1 + x)^3 + x \cdot 3(1 + x)^{3 - 1} \cdot (0 + x^{1 - 1})[/tex]
- Simplify: [tex]\displaystyle y' = (1 + x)^3 + 3x(1 + x)^2[/tex]
- Factor: [tex]\displaystyle y' = (1 + x)^2 \bigg[ (1 + x) + 3x \bigg][/tex]
- Combine like terms: [tex]\displaystyle y' = (1 + x)^2(4x + 1)[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e
