## Calculus: Early Transcendentals 8th Edition

$f_{xyy}=f_{yxy}=f_{yyx}$
Here, we have $f_{xyy}=(f_{xy})_{y}$ and $f_{yxy}=(f_{yx})_{y}$ Further, $(f_{xy})_{y}=(f_{xy})_{y}=f_{xyy}$ This implies that $f_{yxy}=f_{xyy}$ Since, $f_{yyx}=(f_{y})_{yx}$ and $(f_{y})_{yx}$=$(f_{y})_{xy}$ we have $f_{yyx}=(f_{y})_{xy}=f_{yxy}$ This gives: $f_{yxy}=f_{yxy},$ As per the transitive property of equality, we get $f_{xyy}=f_{yxy}=f_{yyx}$