Product and quotient rule ,differentiate each function with respect to x, in calculus

1 answer:

7 0

Answer:

$\displaystyle y' = -100x^4(2x^5 - 1)$

General Formulas and Concepts:

Pre-Algebra

Algebra I

Calculus

Derivatives

Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

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:

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = (-4x^5 + 4)5x^5$

Step 2: Differentiate

Product Rule: $\displaystyle y' = \frac{d}{dx}[(-4x^5 + 4)]5x^5 + (-4x^5 + 4)\frac{d}{dx}[5x^5]$
Basic Power Rule [Derivative Property - Addition/Subtraction]: $\displaystyle y' = (5 \cdot -4x^{5 - 1} + 0)5x^5 + (-4x^5 + 4)(5 \cdot 5x^{5 - 1})$
Simplify: $\displaystyle y' = (-20x^4)5x^5 + (-4x^5 + 4)(25x^4)$
Factor: $\displaystyle y' = 5x^4 \bigg[ (-20x^4)x + (-4x^5 + 4)5 \bigg]$
[Distributive Property] Distributive parenthesis: $\displaystyle y' = 5x^4 \bigg[ -20x^5 - 20x^5 + 20 \bigg]$
Combine like terms: $\displaystyle y' = 5x^4(-40x^5 + 20)$
Factor: $\displaystyle y' = -100x^4(2x^5 - 1)$