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For the relation x²y - y² = 1 - x, find its first derivative at the point where x = 1 and y = 0. Select one: a. O b. О с. O d. e. dy -2y)=-1-2xy (x². dy dx -(1,0) = -1 (x²-2y). dy dx (2x - y) dy (1,0) = 1 dx dy dx dy (1 dx = -1 dy dx = (1,0) = 1-xy dy (2x - 2y) dx = -1-2xy 1 2 -(1,0) = dy (x²-y) - 1 dx dy (1,0) = 0 1 2 = 1-x

For the relation x²y - y² = 1 - x, find its first derivative at the point where x = 1 and y = 0. Select one: a. O b. О с. O d. e. dy -2y)=-1-2xy (x². dy dx -(1,0) = -1 (x²-2y). dy dx (2x - y) dy (1,0) = 1 dx dy dx dy (1 dx = -1 dy dx = (1,0) = 1-xy dy (2x - 2y) dx = -1-2xy 1 2 -(1,0) = dy (x²-y) - 1 dx dy (1,0) = 0 1 2 = 1-x

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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For the relation x²y - y² = 1 - x, find its first derivative at the point where x = 1 and y = 0. Select one: a. O b. O d. gi dy (x²-2y) dx dx dy dx (x²-2y) dy dx (2x - y) dy dx (1,0) = -1 (1,0) = 1 dy dx dy dx (x²-y) dy dx = dy dx = -1 = -(1,0) dy dx -1- 2xy 1 (1,0) = ²/ 2 (2x - 2y) = -1- 2xy dx 1- xy = 1 2 = 1-x (1,0) = 0
Transcribed Image Text:For the relation x²y - y² = 1 - x, find its first derivative at the point where x = 1 and y = 0. Select one: a. O b. O d. gi dy (x²-2y) dx dx dy dx (x²-2y) dy dx (2x - y) dy dx (1,0) = -1 (1,0) = 1 dy dx dy dx (x²-y) dy dx = dy dx = -1 = -(1,0) dy dx -1- 2xy 1 (1,0) = ²/ 2 (2x - 2y) = -1- 2xy dx 1- xy = 1 2 = 1-x (1,0) = 0
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