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: 72

Answer

$\frac{d^2y}{dx^2} = \frac{2}{(x^2+2)^{\frac{3}{2}}}$

Work Step by Step

$y = \sqrt {x^2+2} = (x^2+2)^{\frac{1}{2}}$ Chain Rule: $\frac{dy}{dx} = \frac{1}{2}(x^2+2)^{-\frac{1}{2}}(2x) = x(x^2+2)^{-\frac{1}{2}}$ Product Rule and Chain Rule: $\frac{d^2y}{dx^2} = (x^2+2)^{-\frac{1}{2}} + (x)(-\frac{1}{2})(x^2+2)^{-\frac{3}{2}}(2x) = (x^2+2)^{-\frac{1}{2}}-x^2(x^2+2)^{-\frac{3}{2}} = \frac{2}{(x^2+2)^{\frac{3}{2}}}$
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