Calculus with Applications (10th Edition)

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

Chapter 13 - The Trigonometric Functions - Chapter Review - Review Exercises - Page 701: 55

Answer

$$\frac{{dy}}{{dx}} = 64x{\sin ^3}4{x^2}\cos 4{x^2}$$

Work Step by Step

$$\eqalign{ & y = 2{\sin ^4}\left( {4{x^2}} \right) \cr & {\text{we can write the function as}} \cr & y = 2{\left( {\sin 4{x^2}} \right)^4} \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {2{{\left( {\sin 4{x^2}} \right)}^4}} \right] \cr & {\text{multiple constant rule}} \cr & \frac{{dy}}{{dx}} = 2\frac{d}{{dx}}\left[ {{{\left( {\sin 4{x^2}} \right)}^4}} \right] \cr & {\text{use the general power rule for derivatives }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{dx}}.{\text{ consider }}u = \sin 4{x^2} \cr & then \cr & \frac{{dy}}{{dx}} = 2\left( 4 \right){\left( {\sin 4{x^2}} \right)^{4 - 1}}\frac{d}{{dx}}\left[ {\sin 4{x^2}} \right] \cr & {\text{use }}{D_x}\left( {\sin u} \right) = \cos u \cdot {D_x}\left( u \right) \cr & \frac{{dy}}{{dx}} = 2\left( 4 \right){\left( {\sin 4{x^2}} \right)^3}\left( {\cos 4{x^2}} \right)\frac{d}{{dx}}\left[ {4{x^2}} \right] \cr & \frac{{dy}}{{dx}} = 2\left( 4 \right){\left( {\sin 4{x^2}} \right)^3}\left( {\cos 4{x^2}} \right)\left( {8x} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = 64x{\left( {\sin 4{x^2}} \right)^3}\left( {\cos 4{x^2}} \right) \cr & \frac{{dy}}{{dx}} = 64x{\sin ^3}4{x^2}\cos 4{x^2} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.