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1. Suppose that y is defined implicitly as a function of r by the following equation: 3y² + 2xy + x² = 6 Find an expressions for dy/dr and dy/d²r when e=y=1. Compute the tangent line to the function at (1,1). bydy +2y + 2xdy + 2x = 0 die dy (by +2x) = -24-2x du dy dx = -2y-22 бутах -4 -I when x=y=1 =-(-1/2 2 6/dy)² + byd³y +zdy + 2dy + 2x. 2xdy (dx) dx² dx dx. + 2 = 0 dze² √zy (by +2x) = -2dy - 2 - 6616) -2 day टयू -2dy d2² dx 12

1. Suppose that y is defined implicitly as a function of r by the following equation: 3y² + 2xy + x² = 6 Find an expressions for dy/dr and dy/d²r when e=y=1. Compute the tangent line to the function at (1,1). bydy +2y + 2xdy + 2x = 0 die dy (by +2x) = -24-2x du dy dx = -2y-22 бутах -4 -I when x=y=1 =-(-1/2 2 6/dy)² + byd³y +zdy + 2dy + 2x. 2xdy (dx) dx² dx dx. + 2 = 0 dze² √zy (by +2x) = -2dy - 2 - 6616) -2 day टयू -2dy d2² dx 12

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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1. Suppose that y is defined implicitly as a function of r by the following equation: 3у2 + 2xy + 2 = 6 Find an expressions for dy/d.r and d²y/d².x when & = y = 1. Compute the tangent line to the function at (1,1). bydy dr by due +2y +2xdyne dy +2y +2xdy +2x = 0 dy dx 2 b/dy) 4 (by +2x) = -2y-2x = тву -гу-гх бутах бух (by+2х) dz² dry d'y trdy +2dy +2xdy +2=0 dx² dx dx dze² d2c² 2 – -4. dy du }} 8 =-zdy_2-bdy -zdy dx dx 2-b(dy) =-2-2 bytrx 6 -22 32 - 8 I whe хун 2 - 4 =-4 Wвих чу=1
Transcribed Image Text:1. Suppose that y is defined implicitly as a function of r by the following equation: 3у2 + 2xy + 2 = 6 Find an expressions for dy/d.r and d²y/d².x when & = y = 1. Compute the tangent line to the function at (1,1). bydy dr by due +2y +2xdyne dy +2y +2xdy +2x = 0 dy dx 2 b/dy) 4 (by +2x) = -2y-2x = тву -гу-гх бутах бух (by+2х) dz² dry d'y trdy +2dy +2xdy +2=0 dx² dx dx dze² d2c² 2 – -4. dy du }} 8 =-zdy_2-bdy -zdy dx dx 2-b(dy) =-2-2 bytrx 6 -22 32 - 8 I whe хун 2 - 4 =-4 Wвих чу=1
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