Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 28

Answer

See image. a) $f'(x)=(e^x)\frac{1}{dx}\cos(x)+e^x(\cos(x))\frac{1}{dx}$ $=e^x\cos(x)-e^x\sin(x)$ $f''(x)=e^x\cos(x)-e^x\sin(x)-e^x\sin(x)-e^x\cos(x)=-2e^x\sin(x)$ b) The x value of the maximum point on $f(x)$ is an x intercept of $f'(x)$ and the x value of the maximum point on $f'(x)$ is an x intercept of $f''(x)$, so the derivatives are reasonable

Work Step by Step

a) $f'(x)=(e^x)\frac{1}{dx}\cos(x)+e^x(\cos(x))\frac{1}{dx}$ $=e^x\cos(x)-e^x\sin(x)$ $f''(x)=e^x\cos(x)-e^x\sin(x)-e^x\sin(x)-e^x\cos(x)=-2e^x\sin(x)$ b) The x value of the maximum point on $f(x)$ is an x intercept of $f'(x)$ and the x value of the maximum point on $f'(x)$ is an x intercept of $f''(x)$, so the derivatives are reasonable
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