Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 925: 57

Answer

$v_{ss}=-sin(s^2-t^2)\times 4s^2 +2cos(s^2-t^2)$ $v_{tt}=-sin(s^2-t^2)\times4t^2-2cos(s^2-t^2)$ $v_{st}=v_{ts}=4st\times sin(s^2-t^2)$

Work Step by Step

Take the first partial derivatives of the given function. When taking partial derivative with respect to s, treat t as a constant, and vice versa: $v_{s}=cos(s^2-t^2)\times2s$ $v_{t}=cos(s^2-t^2)\times-2t$ Then take the derivative of the first order partial derivatives to find second partial derivatives: $v_{ss}=-sin(s^2-t^2)\times 4s^2 +2cos(s^2-t^2)$ $v_{tt}=-sin(s^2-t^2)\times4t^2-2cos(s^2-t^2)$ Second partial derivatives of first order partial derivative of s with respect to t and t with respect to s are the same: $v_{st}=v_{ts}=4st\times sin(s^2-t^2)$
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