Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Problems Plus - Problems - Page 433: 5

Answer

$f'(\pi/2)=-1$

Work Step by Step

Find $g'(x)$ using the Chain Rule and the Fundamental Theorem of Calculus in Part 1: $g'(x)=\frac{d}{dx}\int_0^{\cos x}[1+\sin(t^2)]dt$ $g'(x)=\frac{d}{d(\cos x)}\int_0^{\cos x}[1+\sin(t^2)]dt\cdot \frac{d(\cos x)}{dx}$ $g'(x)=[1+\sin(\cos^2 x)]\cdot (-\sin x)$ $g'(x)=-[1+\sin(\cos^2 x)]\sin x$ Find $f'(x)$ using the Chain Rule and the Fundamental Theorem of Calculus in Part 1: $f'(x)=\frac{d}{dx}\int_0^{g(x)}\frac{1}{\sqrt{1+t^3}}dt$ $f'(x)=\frac{d}{d(g(x))}\int_0^{g(x)}\frac{1}{\sqrt{1+t^3}}dt\cdot \frac{d(g(x))}{dx}$ $f'(x)=\frac{1}{1+[g(x)]^3}\cdot \left(-[1+\sin(\cos^2 x)]\sin x\right)$ Find $g(\pi/2)$: $g(\pi/2)=\int_0^{\cos (\pi/2)}[1+\sin((\pi/2)^2)]dt=\int_0^0[1+\sin(\pi^2/4)]dt=0$ Find $f'(\pi/2)$: $f'(\pi/2)=\frac{1}{1+[g(\pi/2)]^3}\cdot \left(-[1+\sin(\cos^2 (\pi/2))]\sin \pi/2\right)$ $f'(\pi/2)=\frac{1}{1+0^3}\cdot \left(-[1+\sin 0]\cdot 1\right)$ $f'(\pi/2)=\frac{1}{1}\cdot \left(-[1+0]\cdot 1\right)$ $f'(\pi/2)=1\cdot -1$ $f'(\pi/2)=-1$
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