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: 73

Answer

$\dfrac{d^2y}{dx^2} = 16x^2e^{-2x^2}-4x^{-2x^2}$

Work Step by Step

$y = e^{-2x^2}$ Chain Rule: $\dfrac{dy}{dx} = -4xe^{-2x^2}$ Product Rule and Chain Rule: $\dfrac{d^2y}{dx^2} = -4e^{-2x^2}+-4x(-4x)(e^{-2x^2}) = 16x^2e^{-2x^2}-4x^{-2x^2}$
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