Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises - Page 423: 48


$g'(x) = \frac{1-(sin~x)^2}{1+(sin~x)^4}\cdot cos~x=\frac{\cos ^3x}{1+(\sin x)^4}$

Work Step by Step

The function $f(t) = \frac{1-t^2}{1+t^4}$ is continuous for all $t$ Let $h(x) = sin~x$ Then $h'(x) = cos~x$ $h'(x)$ is continuous for all values of $x$ According to the Substitution Rule: $g(x) = \int_{1}^{sin~x} \frac{1-t^2}{1+t^4}~dt = \int_{\pi/2}^{x} \frac{1-(sin~u)^2}{1+(sin~u)^4}\cdot cos~u~du$ The function $~~\frac{1-(sin~u)^2}{1+(sin~u)^4}\cdot cos~u~~$ is continuous for all $u$ According to the Fundamental Theorem of Calculus (Part 1): $g'(x) = \frac{1-(sin~x)^2}{1+(sin~x)^4}\cdot cos~x$
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