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Find the derivative of y = (4x³) (5x² − 8) ³ d = 12x² (5x² - 8) *² dy dx ○ du = (127²) (10x² - 8) + dy dx ○ d = (12a²) (5a² − 8) ³ + ½ (40x¹) (5x² − 8) * dy dx (12x²) (10x² - 8)

Find the derivative of y = (4x³) (5x² − 8) ³ d = 12x² (5x² - 8) *² dy dx ○ du = (127²) (10x² - 8) + dy dx ○ d = (12a²) (5a² − 8) ³ + ½ (40x¹) (5x² − 8) * dy dx (12x²) (10x² - 8)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Find the derivative of \( y = \left( 4x^3 \right) \left( 5x^2 - 8 \right)^{\frac{1}{3}} \)** **Options:** - \(\circ \quad \frac{dy}{dx} = 12x^2 \left( 5x^2 - 8 \right)^{-\frac{2}{3}}\) - \(\circ \quad \frac{dy}{dx} = \left( 12x^2 \right) \left( 10x^2 - 8 \right)^{-\frac{2}{3}}\) - \(\circ \quad \frac{dy}{dx} = \left( 12x^2 \right) \left( 5x^2 - 8 \right)^{\frac{1}{3}} + \frac{1}{3} \left( 40x^4 \right) \left( 5x^2 - 8 \right)^{-\frac{2}{3}}\) - \(\circ \quad \left( 12x^2 \right) (10x^2 - 8)\) **Explanation:** This problem involves finding the derivative of a function defined as a product of two terms. The expression inside involves a power term and a polynomial. You would typically apply the product rule and the chain rule to find the derivative. Each option presents a different form of expressing the derivative.
Transcribed Image Text:**Find the derivative of \( y = \left( 4x^3 \right) \left( 5x^2 - 8 \right)^{\frac{1}{3}} \)** **Options:** - \(\circ \quad \frac{dy}{dx} = 12x^2 \left( 5x^2 - 8 \right)^{-\frac{2}{3}}\) - \(\circ \quad \frac{dy}{dx} = \left( 12x^2 \right) \left( 10x^2 - 8 \right)^{-\frac{2}{3}}\) - \(\circ \quad \frac{dy}{dx} = \left( 12x^2 \right) \left( 5x^2 - 8 \right)^{\frac{1}{3}} + \frac{1}{3} \left( 40x^4 \right) \left( 5x^2 - 8 \right)^{-\frac{2}{3}}\) - \(\circ \quad \left( 12x^2 \right) (10x^2 - 8)\) **Explanation:** This problem involves finding the derivative of a function defined as a product of two terms. The expression inside involves a power term and a polynomial. You would typically apply the product rule and the chain rule to find the derivative. Each option presents a different form of expressing the derivative.
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