Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 151: 51


$-\cos x$

Work Step by Step

$\frac{d}{dx}(\sin x)=\cos x$ $\frac{d^{2}}{dx^{2}}(\sin x)= \frac{d}{dx}(\cos x)= -\sin x$ $\frac{d^{3}}{dx^{3}}(\sin x)=\frac{d}{dx}(-\sin x)= -\cos x$ $\frac{d^{4}}{dx^{4}}(\sin x)=\frac{d}{dx}(-\cos x)=\sin x$ This pattern will continue and we can generalise: $\frac{d^{4n+1}}{dx^{4n+1}}(\sin x)=\cos x$ $\frac{d^{4n+2}}{dx^{4n+2}}(\sin x)=-\sin x$ $\frac{d^{4n+3}}{dx^{4n+3}}(\sin x)=-\cos x$ $\frac{d^{4n+4}}{dx^{4n+4}}(\sin x)=\sin x$ where n=0,1,2,3... Therefore, $\frac{d^{99}}{dx^{99}}(\sin x)=\frac{d^{4\times24+3}}{dx^{4\times24+3}}(\sin x)= -\cos x$
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