Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6 - Page 158: 51



Work Step by Step

$$\frac{dy}{dx}=(x\cos(5x)-\sin^2x)'=x'\cos(5x)+x(\cos(5x))'-2\sin x(\sin x)'= \cos(5x)+x(-\sin(5x))(5x)'-2\sin x\cos x= \cos(5x)-5x\sin(5x)-2\sin x\cos x$$ $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{dy}{dx}\Big)=(\cos(5x)-5x\sin(5x)-2\sin x\cos x)'= -\sin(5x)(5x)'-5((x)'\sin (5x)+x(\sin(5x)')-2((\sin x)'\cos x+\sin x(\cos x)')= -5\sin(5x)-5\sin(5x)-5x\cos(5x)(5x)'-2\cos^2x+2\sin^2x= -10\sin(5x)-25x\cos(5x)-2\cos(2x)$$
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