Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.7 - Implicit Differentiation - Exercises 3.7 - Page 155: 15

Answer

$\frac{2e^{2x}-cos(x+3y)}{3cos(x+3y)}$

Work Step by Step

We start with: $e^{2x}=\sin(x+3y)$ Now use implicit differentiation: $2e^{2x}=(1+3y') \cos(x+3y)$ And solve for $y'$: $1+3y'=\frac{2e^{2x}}{cos(x+3y)}$ $3y'=\frac{2e^{2x}}{cos (x+3y)}-1$ $y'=\frac{2e^{2x}-cos(x+3y)}{3cos(x+3y)}$
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