Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 6 - Applications of the Derivative - 6.4 Implicit Differentiation - 6.4 Exercises: 5

Answer

\[{y^,} = \frac{{15{x^2}}}{{6y + 4}}\]

Work Step by Step

\[\begin{gathered} 5{x^3}\,\, = 3{y^2} + 4y \hfill \\ \,Find\,\,dy/dx\,\,\,by\,\,implicit\,\,differentiation \hfill \\ \frac{d}{{dx}}\,\left( {5{x^3}} \right) = \frac{d}{{dx}}\,\left( {3{y^2} + 4y} \right) \hfill \\ by\,\,the\,\,chain\,\,rule \hfill \\ 15{x^2} = 6y{y^,} + 4{y^,} \hfill \\ 6y{y^,} + 14{y^,} = 15{x^2} \hfill \\ Factor \hfill \\ {y^,}\,\left( {6y + 4} \right) = 15{x^2} \hfill \\ {y^,} = \frac{{15{x^2}}}{{6y + 4}} \hfill \\ Then \hfill \\ {y^,} = \frac{{15{x^2}}}{{6y + 4}} \hfill \\ \end{gathered} \]
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