Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.6 Implicit Differentiation - 2.6 Exercises: 13

Answer

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

Work Step by Step

$\sqrt {x+y} = x^4 + y^4$ Differentiate both sides with respect to x. $\frac{1+\frac{dy}{dx}}{2\sqrt{x+y}} = 4x^3 + 4y^3\frac{dy}{dx}$ Isolate $\frac{dy}{dx}$ $1+\frac{dy}{dx} = 2\sqrt{x+y}(4x^3 + 4y^3\frac{dy}{dx})$ $\frac{dy}{dx} - 8y^3\sqrt{x+y}\frac{dy}{dx} = 8x^3\sqrt{x+y}-1$ $\frac{dy}{dx}(1-8y^3\sqrt{x+y}) = 8x^3\sqrt{x+y}-1$ $\frac{dy}{dx}=\frac{8x^3\sqrt{x+y}-1}{1-8y^3\sqrt{x+y}}$
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