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: 20

Answer

\[\frac{{dy}}{{dx}} = \frac{{1 - 3y\,{{\left( {xy + 1} \right)}^2}}}{{3x\,{{\left( {xy + 1} \right)}^2} + 2y}}\]

Work Step by Step

\[\begin{gathered} \,{\left( {xy + 1} \right)^3} = x - {y^2} + 8 \hfill \\ \hfill \\ implicit\,\,differentiation \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {\,{{\left( {xy + 1} \right)}^3}} \right] = \frac{d}{{dx}}\,\,\left[ {x - {y^2} + 8} \right] \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \end{gathered} \] \[3\,{\left( {xy + 1} \right)^2}\,\left( {xy' + y} \right) = 1 - 2yy'\] \[multiply\] \[3xy'\,{\left( {xy + 1} \right)^2} + 3y\,{\left( {xy + 1} \right)^2} = 1 - 2yy'\] \[collect\,\,like\,\,terms\] \[3xy'\,{\left( {xy + 1} \right)^2} + 2yy' = 1 - 3y\,{\left( {xy + 1} \right)^2}\] \[factor\,\,{y^,}\] \[y' = \,\,\left[ {3x\,{{\left( {xy + 1} \right)}^2} + 2y} \right] = 1 - 3y\,{\left( {xy + 1} \right)^2}\] \[solve\,\,for\,\,{y^,}\] \[\frac{{dy}}{{dx}} = \frac{{1 - 3y\,{{\left( {xy + 1} \right)}^2}}}{{3x\,{{\left( {xy + 1} \right)}^2} + 2y}}\]
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