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: 13

Answer

$y' = \frac{1-8x^{3}\sqrt {x+y}}{8y^{3}\sqrt {x+y} -1}$

Work Step by Step

Equation: $\sqrt{x+y} = x^{4} + y^{4}$. Differentiate both sides: $\frac{1+y'}{2\sqrt {x+y}} = 4x^{3} + 4y'y^{3}$. Simplify and solve for y': $1+y' = 8y'y^{3}\sqrt {x+y} + 8x^{3}\sqrt {x+y}$. $1-8x^{3}\sqrt {x+y} = y'(8y^{3}\sqrt {x+y} - 1)$. $y' = \frac{1-8x^{3}\sqrt {x+y}}{8y^{3}\sqrt {x+y} -1}$
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