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: 28

Answer

\[y = - x + 2\]

Work Step by Step

\[\begin{gathered} {x^3} + {y^3} = 2xy\,\,\,\,\,\,,\,\,\,\left( {1,1} \right) \hfill \\ \hfill \\ use\,\,the\,\,implicit\,\,differentiation \hfill \\ \hfill \\ 3{x^2} + 3{y^2}{y^,} = 2x{y^,} + 2y \hfill \\ \hfill \\ collect\,\,like\,\,terms \hfill \\ \hfill \\ 3{y^2}{y^,} - 2x{y^,} = 2y - 3{x^2} \hfill \\ \hfill \\ solve\,\,for\,\,y{\,^,} \hfill \\ \hfill \\ {y^,}\,\left( {3{y^2} - 2x} \right) = 2y - 3{x^2} \hfill \\ \hfill \\ {y^,} = \frac{{2\,\left( 1 \right) - 3\,{{\left( 1 \right)}^2}}}{{3\,{{\left( 1 \right)}^2} - 2\,\left( 1 \right)}} = 1 \hfill \\ \hfill \\ use\,\,the\,\,point\, - \,slope\,\,form \hfill \\ \hfill \\ y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ \hfill \\ then \hfill \\ \hfill \\ y - 1 = - 1\,\left( {x - 1} \right) \hfill \\ \hfill \\ y - 1 = - x + 1 \hfill \\ \hfill \\ y = - x + 2 \hfill \\ \hfill \\ \end{gathered} \]
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