Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises: 13

Answer

\[y' = \frac{{1 - y\cos xy}}{{x\cos xy - 1}}\]

Work Step by Step

\[find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.\] \[\cos xy\, \cdot {\left( {xy} \right)^\prime } = 1 + y'\] \[Using\,\,the\,\,product\,\,rule\] \[\cos xy \cdot \,\left( {y + xy'} \right) = 1 + y'\] \[multiply\] \[\cos xy \cdot xy' - y' = 1 - \cos xy \cdot y\] \[factor\] \[y'\,\left( {x\cos xy - 1} \right) = 1 - y\cos xy\] \[solve\,\,for\,\,{y^{\,,}}\] \[y' = \frac{{1 - y\cos xy}}{{x\cos xy - 1}}\]
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