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: 38

Answer

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

Work Step by Step

\[\begin{gathered} y = \sqrt[3]{{{x^2} - x + 1}} \hfill \\ \hfill \\ rewrite\, \hfill \\ \hfill \\ y = \,{\left( {{x^2} - x + 1} \right)^{\frac{1}{3}}} \hfill \\ \hfill \\ differentiate \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{1}{3}\,{\left( {{x^2} - x + 1} \right)^{ - \frac{2}{3}}}\,\left( {2x - 1} \right) \hfill \\ \hfill \\ use\,\,{x^{ - m}} = \frac{1}{{{x^m}}} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{2x - 1}}{{3\,{{\left( {{x^2} - x + 1} \right)}^{\frac{2}{3}}}}} \hfill \\ \end{gathered} \]
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