Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 215: 31



Work Step by Step

$2(x^2+y^2)^2=25(x^2-y^2)\\ 4(x^2+y^2)(2x+2yy')=25(2x-2yy')\\ 4(x^2+y^2)(x+yy')=25(x-yy')\\ 4x^3+4x^2yy'+4y^2x+4y^3y'=25x-25yy'\\ 4x^2yy'+4y^3y'+25yy'=25x-4x^3-4y^2x\\ y'(4x^2y+4y^3+25y)=25x-4x^3-4y^2x\\ y'=\frac{25x-4x^3-4y^2x}{4x^2y+4y^3+25y}$ Plug in $(3,1)$ into $y′$ to find the gradient $y'=\frac{25(3)-4(3)^3-4(1)^2(3)}{4(3)^2(1)+4(1)^3+25(1)}\\ y'=-\frac{9}{13}$ Equation of the tangent line at $(3,1)$ $y-1=-\frac{9}{13}(x-3)\\ y=-\frac{9}{13}x+\frac{40}{13}$
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