Question

Differentiate the following functions (i) x(1+x)^3​

Answer 1

Answer:

$\displaystyle y' = (1 + x)^2(4x + 1)$

General Formulas and Concepts:

Algebra I

• Terms/Coefficients
• Functions
• Function Notation
• Factoring

Calculus

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]:                                                                $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:                                                                                $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Chain Rule]:                                                                                       $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Explanation:

Step 1: Define

Identify

y = x(1 + x)³

Step 2: Differentiate

1. Product Rule [Derivative Rule - Chain Rule]:                                                  $\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot \frac{d}{dx}[1 + x]$
2. Derivative Property [Addition/Subtraction]:                                                    $\displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot (\frac{d}{dx}[1] + \frac{d}{dx}[x])$
3. Basic Power Rule:                                                                                             $\displaystyle y' = x^{1 - 1} \cdot (1 + x)^3 + x \cdot 3(1 + x)^{3 - 1} \cdot (0 + x^{1 - 1})$
4. Simplify:                                                                                                             $\displaystyle y' = (1 + x)^3 + 3x(1 + x)^2$
5. Factor:                                                                                                               $\displaystyle y' = (1 + x)^2 \bigg[ (1 + x) + 3x \bigg]$
6. Combine like terms:                                                                                         $\displaystyle y' = (1 + x)^2(4x + 1)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e