Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.3 Differentiation Rules - Exercises 3.3 - Page 125: 34

Answer

$$\frac{{ds}}{{dt}} = - \frac{5}{{{t^2}}} + \frac{2}{{{t^3}}}{\text{ and }}\frac{{{d^2}s}}{{d{t^2}}} = \frac{{10}}{{{t^3}}} - \frac{6}{{{t^4}}}$$

Work Step by Step

$$\eqalign{ & s = \frac{{{t^2} + 5t - 1}}{{{t^2}}} \cr & {\text{Use the distributive property}} \cr & s = \frac{{{t^2}}}{{{t^2}}} + \frac{{5t}}{{{t^2}}} - \frac{1}{{{t^2}}} \cr & s = 1 + \frac{5}{t} - \frac{1}{{{t^2}}} \cr & {\text{where }}\frac{1}{{{t^n}}} = {t^{ - n}} \cr & s = 1 + 5{t^{ - 1}} - {t^{ - 2}} \cr & \cr & {\text{Find the first derivative}} \cr & \frac{{ds}}{{dt}} = \frac{d}{{dt}}\left[ {1 + 5{t^{ - 1}} - {t^{ - 2}}} \right] \cr & {\text{use }}\frac{d}{{dt}}\left[ {{s^n}} \right] = n{s^{n - 1}} \cr & \frac{{ds}}{{dt}} = 0 + 5\left( { - {t^{ - 2}}} \right) - \left( { - 2{t^{ - 3}}} \right) \cr & \frac{{ds}}{{dt}} = - 5{t^{ - 2}} + 2{t^{ - 3}} \cr & \frac{{ds}}{{dt}} = - \frac{5}{{{t^2}}} + \frac{2}{{{t^3}}} \cr & \cr & \cr & {\text{Find the second derivative}} \cr & \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left[ { - 5{t^{ - 2}} + 2{t^{ - 3}}} \right] \cr & {\text{use }}\frac{d}{{dt}}\left[ {{s^n}} \right] = n{s^{n - 1}} \cr & \frac{{{d^2}s}}{{d{t^2}}} = - 5\left( { - 2{t^{ - 3}}} \right) + 2\left( {-3{t^{ - 4}}} \right) \cr & \frac{{{d^2}s}}{{d{t^2}}} = 10{t^{ - 3}} - 6{t^{ - 4}} \cr & \frac{{{d^2}s}}{{d{t^2}}} = \frac{{10}}{{{t^3}}} - \frac{6}{{{t^4}}} \cr} $$
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