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 - Page 200: 43

Answer

$$\frac{{dy}}{{dx}} = \frac{2}{{9\root 3 \of {{x^2}\left( {1 + {x^{1/3}}} \right)} }}$$

Work Step by Step

$$\eqalign{ & y = \root 3 \of {{{\left( {1 + {x^{1/3}}} \right)}^2}} \cr & {\text{Write as}} \cr & y = {\left( {{{\left( {1 + {x^{1/3}}} \right)}^2}} \right)^{1/3}} \cr & y = {\left( {1 + {x^{1/3}}} \right)^{2/3}} \cr & {\text{differentiate}} \cr & \frac{{dy}}{{dx}} = \frac{2}{3}{\left( {1 + {x^{1/3}}} \right)^{ - 1/3}}\frac{d}{{dx}}\left[ {1 + {x^{1/3}}} \right] \cr & {\text{Then}} \cr & \frac{{dy}}{{dx}} = \frac{2}{3}{\left( {1 + {x^{1/3}}} \right)^{ - 1/3}}\left( {\frac{1}{3}{x^{ - 2/3}}} \right) \cr & \frac{{dy}}{{dx}} = \frac{2}{{9{x^{2/3}}{{\left( {1 + {x^{1/3}}} \right)}^{1/3}}}} \cr & \frac{{dy}}{{dx}} = \frac{2}{{9{{\left[ {{x^2}\left( {1 + {x^{1/3}}} \right)} \right]}^{1/3}}}} \cr & \frac{{dy}}{{dx}} = \frac{2}{{9\root 3 \of {{x^2}\left( {1 + {x^{1/3}}} \right)} }} \cr} $$
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