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: 57

Answer

$$\frac{d}{d\omega}(a\cos^2\pi\omega+b\sin^2\pi\omega)=(b\pi-a\pi)\sin2\pi\omega$$

Work Step by Step

$$\frac{d}{d\omega}(a\cos^2\pi\omega+b\sin^2\pi\omega)=a\cdot2\cos\pi\omega\frac{d}{d\omega}(\cos\pi\omega)+b\cdot2\sin\pi\omega\frac{d}{d\omega}(\sin\pi\omega)= 2a\cos\pi\omega\cdot(-\sin\pi\omega)\frac{d}{d\omega}(\pi\omega)+2b\sin\pi\omega\cdot\cos\pi\omega\frac{d}{d\omega}(\pi\omega)= -2a\cos\pi\omega\sin\pi\omega\cdot\pi+2b\cos\pi\omega\sin\pi\omega\cdot\pi= (b\pi-a\pi)\cdot 2\cos\pi\omega\sin\pi\omega=(b\pi-a\pi)\sin2\pi\omega$$
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