Answer
$f_{xx}(x, y)=6xy^{5}+24x^{2}y$
$f_{xy}(x, y)=15x^{2}y^{4}+8x^{3}$
$f_{yx}(x, y) =15x^{2}y^{4}+8x^{3}$
$f_{yy}(x, y)=20x^{3}y^{3}$.
Work Step by Step
$f(x, y)=x^{3}y^{5}+2x^{4}y$
$f_{x}(x, y)=\displaystyle \frac{\partial}{\partial x}[f(x, y)]=(3x^{2})y^{5}+2(4x^{3})y=3x^{2}y^{5}+8x^{3}y$,
$f_{xx}(x, y)=\displaystyle \frac{\partial}{\partial x}[f_{x}(x, y)]=3(2x)y^{5}+8(3x^{2})y=6xy^{5}+24x^{2}y$,
$f_{xy}(x, y)=\displaystyle \frac{\partial}{\partial y}[f_{x}(x, y)]=3x^{2}(5y^{4})+8x^{3}(1)=15x^{2}y^{4}+8x^{3}$
$f_{y}(x, y)=\displaystyle \frac{\partial}{\partial y}[f(x, y)]=x^{3}(5y^{4})+2x^{4}(1) =5x^{3}y^{4}+2x^{4}$.
$f_{yx}(x, y) =\displaystyle \frac{\partial}{\partial x}[f_{y}(x, y)]=15x^{2}y^{4}+8x^{3}$
$f_{yy}(x, y)=\displaystyle \frac{\partial}{\partial y}[f_{y}(x, y)]=20x^{3}y^{3}$.
