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: 35

Answer

\[{y^{,,}} = \frac{{4{e^{2y}}}}{{\,\left( {1 - 2{e^{2y}}} \right)^3}}\]

Work Step by Step

\[\begin{gathered} {e^{2y}} + x = y \hfill \\ \hfill \\ use\,\,the\,\,Implicit\,\,differentiation \hfill \\ \hfill \\ 2{y^,}\,{e^{2y}} + 1 = {y^,} \hfill \\ \hfill \\ factor\,\,{y^,} \hfill \\ \hfill \\ {y^,} = \,\left( {2{e^{2y}} - 1} \right) = - 1 \hfill \\ \hfill \\ solve\,\,for\,\,{y^,} \hfill \\ \hfill \\ {y^,} = \frac{1}{{1 - 2{e^{2y}}}} \hfill \\ \hfill \\ Implicit\,\,differentiation \hfill \\ \hfill \\ {y^{,,}} = \frac{{ - 4{y^,}{e^{2y}}}}{{\,{{\left( {1 - 2{e^{2y}}} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {y^{,,}} = \frac{{4\,\left( {\frac{1}{{1 - 2{e^{2y}}}}} \right){e^{2y}}}}{{\,\left( {1 - 2{e^{2y}}} \right)^2}} \hfill \\ \hfill \\ {y^{,,}} = \frac{{4{e^{2y}}}}{{\,\left( {1 - 2{e^{2y}}} \right)^3}} \hfill \\ \hfill \\ \end{gathered} \]
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