Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 782: 63

Answer

${f_{xyxzy}} = 0$

Work Step by Step

Using Clairaut's Theorem, we arrange the order of the partial differentiations such that: ${f_{xyxzy}} = {f_{xxyyz}}$ So, ${f_x} = yz\cos \left( {xz} \right)\sin \left( {x + z} \right) + y\sin \left( {xz} \right)\cos \left( {x + z} \right) + \tan y + \tan \left( {\frac{{z + {z^{ - 1}}}}{{y - {y^{ - 1}}}}} \right)$ ${f_{xx}} = - y{z^2}\sin \left( {xz} \right)\sin \left( {x + z} \right) + yz\cos \left( {xz} \right)\cos \left( {x + z} \right)$ $ + yz\cos \left( {xz} \right)\cos \left( {x + z} \right) - y\sin \left( {xz} \right)\sin \left( {x + z} \right)$ ${f_{xxy}} = - {z^2}\sin \left( {xz} \right)\sin \left( {x + z} \right) + z\cos \left( {xz} \right)\cos \left( {x + z} \right)$ $ + z\cos \left( {xz} \right)\cos \left( {x + z} \right) - \sin \left( {xz} \right)\sin \left( {x + z} \right)$ ${f_{xxyy}} = 0$ ${f_{xxyyz}} = 0$ Therefore, ${f_{xyxzy}} = {f_{xxyyz}} = 0$
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