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: 20

Answer

$dy/dx=\frac{(1+x^2)^2(sec(x-y))^2+2xy}{(1+x^2)^2(sec(x-y))^2+x^2+1}$

Work Step by Step

Take the derivative as is on either side of the equation: $(sec(x-y))^2\times(1-dy/dx)=\frac{(dy/dx)x^2+dy/dx-2xy}{(1+x^2)^2}$ Move all terms with dy/dx onto one side of the equal sign and distribute the dy/dx out of each term: $dy/dx((1+x^2)^2(sec(x-y))^2+x^2+1)=(1+x^2)^2(sec(x-y))^2+2xy$ Isolate dy/dx by dividing both sides by the terms: $dy/dx=\frac{(1+x^2)^2(sec(x-y))^2+2xy}{(1+x^2)^2(sec(x-y))^2+x^2+1}$
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