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: 17

Answer

\[y' = \frac{1}{{{e^y} + 2y\sin {y^2}}}\]

Work Step by Step

\[find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.\] \[ - \sin \,{y^2} \cdot \,\,{\left( {{y^2}} \right)^\prime } + 1 = {e^y} \cdot y'\] \[ - \sin \,\,{y^2} \cdot \,\,{\left( {{y^2}} \right)^\prime } + 1 = {e^y} \cdot y'\] \[differentiate\] \[y'{e^y} + 2yy'\sin {y^2} = 1\] \[factor\,\,out\,\,y'\] \[y'\,\left( {{e^y} + 2y\sin {y^2}} \right) = 1\] \[solve\,\,for\,\,y'\] \[y' = \frac{1}{{{e^y} + 2y\sin {y^2}}}\]
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