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: 6

Answer

\[\frac{{dy}}{{dx}} = \frac{{9{x^2}}}{{16y + 10}}\]

Work Step by Step

\[\begin{gathered} 3{x^3} - 8{y^2} = 10y \hfill \\ \,Find\,\,dy/dx\,\,\,by\,\,implicit\,\,differentiation \hfill \\ \frac{d}{{dx}}\,\left( {3{x^3} - 8{y^2}} \right) = \frac{d}{{dx}}\,\left( {10y} \right) \hfill \\ By\,\,the\,\,chain\,\,rule \hfill \\ 9{x^2} - 8\,\left( 2 \right)y{y^,} = 10{y^,} \hfill \\ 9{x^2} - 16y{y^,} = 10{y^,} \hfill \\ Move\,\,all\,\,{y^,}\,\,terms\,\,to\,\,the\,\,same \hfill \\ - 16y{y^,} - 10{y^,} = - 9{x^2} \hfill \\ Factor \hfill \\ {y^,}\,\left( {16y + 10} \right) = 9{x^2} \hfill \\ {y^,} = \frac{{9{x^2}}}{{16y + 10}} \hfill \\ Then \hfill \\ \frac{{dy}}{{dx}} = \frac{{9{x^2}}}{{16y + 10}} \hfill \\ \end{gathered} \]
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