Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises - Page 244: 12

Answer

$$\frac{{dy}}{{dx}} = 21{x^2} - 8x - 5$$

Work Step by Step

$$\eqalign{ & y = 7{x^3} - 4{x^2} - 5x + \sqrt 2 \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {7{x^3} - 4{x^2} - 5x + \sqrt 2 } \right] \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {7{x^3}} \right] - \frac{d}{{dx}}\left[ {4{x^2}} \right] - \frac{d}{{dx}}\left[ {5x} \right] + \frac{d}{{dx}}\left[ {\sqrt 2 } \right] \cr & {\text{use the power rule for differentiation }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}}{\text{ and }}\frac{d}{{dx}}\left[ k \right] = 0 \cr & \frac{{dy}}{{dx}} = 21{x^2} - 8x - 5 + 0 \cr & \frac{{dy}}{{dx}} = 21{x^2} - 8x - 5 \cr} $$
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