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: 42

Answer

\[{y^,} = \,{\left( {x + 1} \right)^{\frac{1}{3}}} + \frac{x}{{3\,{{\left( {x + 1} \right)}^{\frac{2}{3}}}}}\]

Work Step by Step

\[\begin{gathered} y = x\,{\left( {x + 1} \right)^{\frac{1}{3}}} \hfill \\ \hfill \\ use\,\,the\,\,product\,\,rule \hfill \\ \hfill \\ {y^,} = {x^,} \cdot \,{\left( {x + 1} \right)^{\frac{1}{3}}} + x \cdot \,{\left( {\,{{\left( {x + 1} \right)}^{\frac{1}{3}}}} \right)^,} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {y^,} = \,{\left( {x + 1} \right)^{\frac{1}{3}}} + x \cdot \frac{1}{3}\,{\left( {x + 1} \right)^{ - \frac{2}{3}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {y^,} = \,{\left( {x + 1} \right)^{\frac{1}{3}}} + \frac{x}{{3\,{{\left( {x + 1} \right)}^{\frac{2}{3}}}}} \hfill \\ \hfill \\ \end{gathered} \]
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