## Calculus: Early Transcendentals 8th Edition

$f_{xzy}=f_{yxz}=6yz^2$.
$f(x,y,z)=xy^2z^3+\arcsin{(x\sqrt z)}$ In order to find $f_{xzy}$ we can differentiate in whatever order we want because of Clairaut's Theorem. $f_y=2xyz^3\Rightarrow f_{yx}=2yz^3\Rightarrow f_{yxz}=6yz^2$