Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.7 - Implicit Differentiation - Exercises 3.7 - Page 155: 16

Answer

$$\frac{{dr}}{{d\theta }} = \frac{1}{{\root 3 \of \theta }} + \frac{1}{{\root 4 \of \theta }} + \frac{1}{{\sqrt \theta }}$$

Work Step by Step

$$\eqalign{ & r - 2\sqrt \theta = \frac{3}{2}{\theta ^{2/3}} + \frac{4}{3}{\theta ^{3/4}} \cr & {\text{differentiate both sides with respect to }}\theta \cr & \frac{d}{{d\theta }}\left( r \right) - \frac{d}{{d\theta }}\left( {2\sqrt \theta } \right) = \frac{d}{{d\theta }}\left( {\frac{3}{2}{\theta ^{2/3}}} \right) + \frac{d}{{d\theta }}\left( {\frac{4}{3}{\theta ^{3/4}}} \right) \cr & \cr & {\text{Find derivatives}} \cr & \frac{{dr}}{{d\theta }} - 2\left( {\frac{1}{{2\sqrt \theta }}} \right) = \frac{3}{2}\left( {\frac{2}{3}{\theta ^{ - 1/3}}} \right) + \frac{4}{3}\frac{d}{{d\theta }}\left( {\frac{3}{4}{\theta ^{ - 1/4}}} \right) \cr & \frac{{dr}}{{d\theta }} - \frac{1}{{\sqrt \theta }} = {\theta ^{ - 1/3}} + {\theta ^{ - 1/4}} \cr & \cr & {\text{Solve for }}\frac{{dr}}{{d\theta }} \cr & \frac{{dr}}{{d\theta }} = {\theta ^{ - 1/3}} + {\theta ^{ - 1/4}} + \frac{1}{{\sqrt \theta }} \cr & \frac{{dr}}{{d\theta }} = \frac{1}{{\root 3 \of \theta }} + \frac{1}{{\root 4 \of \theta }} + \frac{1}{{\sqrt \theta }} \cr} $$
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