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Find dy/dx at the given point for the equation. x = y3 – 5y2 + 2, (-2,1)

Find dy/dx at the given point for the equation. x = y3 – 5y2 + 2, (-2,1)

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 dy/dx at the given point for the equation. x = y3 – 5y2 + 2, (-2, 1) Step 1 The given function is x = y³ − 5y² + 2. Differentiate the given function with respect to x. d dx d -X = dx 1 = 1 = 1 = dy dx || -(y³ - 5y² + 2) d d (y³) - 50 (y²) + (2) 5. dx dx dx dy dx 1 dy dx 10y dy dx - 10y
Transcribed Image Text:Find dy/dx at the given point for the equation. x = y3 – 5y2 + 2, (-2, 1) Step 1 The given function is x = y³ − 5y² + 2. Differentiate the given function with respect to x. d dx d -X = dx 1 = 1 = 1 = dy dx || -(y³ - 5y² + 2) d d (y³) - 50 (y²) + (2) 5. dx dx dx dy dx 1 dy dx 10y dy dx - 10y
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