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A heart-shaped graph in a 2D plane is defined by the equation (x² + y² - 1)³= 4x²y³ Determine dy/dx by implicit differentiation. O y = 3 . O y' y 3(x²+y²-1)²-4y³ (x²+y²-1)²-2r²y = 62²y²-3y(x² + y² −1)² 3x(x²+y²-1)²-4xy³ O y'= 6x(x² + y² - 1)² - 8xy³ O y' = 6x(x²+y²-1)² +8ry³ 12x²y²-6y(x²+y²-1)²

A heart-shaped graph in a 2D plane is defined by the equation (x² + y² - 1)³= 4x²y³ Determine dy/dx by implicit differentiation. O y = 3 . O y' y 3(x²+y²-1)²-4y³ (x²+y²-1)²-2r²y = 62²y²-3y(x² + y² −1)² 3x(x²+y²-1)²-4xy³ O y'= 6x(x² + y² - 1)² - 8xy³ O y' = 6x(x²+y²-1)² +8ry³ 12x²y²-6y(x²+y²-1)²

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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Question 14 A heart-shaped graph in a 2D plane is defined by the equation (x² + y² - 1)³ = 4x²y³ Determine dy/dx by implicit differentiation. y' = O -x 3y . 6r²y²-3y(r²+y²-1)² 3x(x² + y²-1)²-4xy³ O y = 6x(x² + y² - 1)² - 8xy³ Oy' = < Previous 3(x² + y²-1)²-4y³ (x² + y²-1)²-2r²y 6x(x²+y²-1)² +8ry³ 12x²y²-6y(x² + y²-1)²
Transcribed Image Text:Question 14 A heart-shaped graph in a 2D plane is defined by the equation (x² + y² - 1)³ = 4x²y³ Determine dy/dx by implicit differentiation. y' = O -x 3y . 6r²y²-3y(r²+y²-1)² 3x(x² + y²-1)²-4xy³ O y = 6x(x² + y² - 1)² - 8xy³ Oy' = < Previous 3(x² + y²-1)²-4y³ (x² + y²-1)²-2r²y 6x(x²+y²-1)² +8ry³ 12x²y²-6y(x² + y²-1)²
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