Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises: 67

Answer

$y'= (2p+2\pi)(\sin(p^2)) + (p+\pi)^2(2p\cos(p^2))$

Work Step by Step

Chain Rule $\frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x)$ $y=(p+\pi)^2sin(p^2)$ Product Rule: $y' = (\frac{d}{dx}(p+\pi)^2)(sin(p^2))+((p+\pi)^2)(\frac{d}{dx}sin(p^2))$ Chain Rule: $y'=(2(p+\pi))(sin(p^2))+((p+\pi)^2)(2pcos(p^2)) = (2p+2\pi)(\sin(p^2)) + (p+\pi)^2(2p\cos(p^2))$
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