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: 19

Answer

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

Work Step by Step

\[\begin{gathered} {x^3} = \frac{{x + y}}{{x - y}} \hfill \\ \hfill \\ implicit\,\,differentiation \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {{x^3}} \right] = \frac{d}{{dx}}\,\,\left[ {\frac{{x + y}}{{x - y}}} \right] \hfill \\ \hfill \\ then \hfill \\ \hfill \\ 3{x^2} = \frac{{\,\left( {x - y} \right)\,\left( {1 + y'} \right) - \,\left( {x + y} \right)\,\left( {1 - y'} \right)}}{{\,{{\left( {x - y} \right)}^2}}} \hfill \\ \hfill \\ mutliply \hfill \\ \hfill \\ 3{x^2} = \frac{{x + xy' - y - yy' - x + xy' - y + yy'}}{{\,{{\left( {x - y} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ 3{x^2} = \frac{{2xy' - 2y}}{{\,{{\left( {x - y} \right)}^2}}} \hfill \\ \hfill \\ solve\,\,for\,\,{y^,} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{3{x^2}\,{{\left( {x - y} \right)}^2} + 2y}}{{2x}} \hfill \\ \end{gathered} \]
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