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: 21

Answer

\[\frac{{dy}}{{dx}} = \frac{{13y - 18{x^2}}}{{21{y^2} - 13x}}\]

Work Step by Step

\[\begin{gathered} 6{x^3} + 7{y^3} = 13xy \hfill \\ \hfill \\ implicit\,\,differentiation \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {6{x^3} + 7{y^3}} \right] = \frac{d}{{dx}}\,\,\left[ {13xy} \right] \hfill \\ \hfill \\ use\,\,product\,\,rule \hfill \\ \hfill \\ 18{x^2} + 21{y^2}y' = 13xy' + 13y \hfill \\ \hfill \\ collect\,\,like\,\,terms \hfill \\ \hfill \\ 21{y^2}y' - 13xy' = 13y - 18{x^2} \hfill \\ \hfill \\ solve\,\,for\,\,{y^,} \hfill \\ \hfill \\ y'\,\left( {21{y^2} - 13x} \right) = 13y - 18{x^2} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{13y - 18{x^2}}}{{21{y^2} - 13x}} \hfill \\ \hfill \\ \end{gathered} \]
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