## Calculus with Applications (10th Edition)

a. $\frac{\partial z}{\partial y}=f_{yy}(x,y)=\frac{4x^{3}+8x^{2}y^{2}-12xy^{4}}{(x-y^{2})^{4}}$ $\frac{\partial z}{\partial y}=f_{yx}(x,y)=\frac{4y^{5}-8x^{2}y+4x^{2}y}{(x-y^{2})^{4}}$ b. $\frac{\partial z}{\partial x}=f_{xx}(0,2)=\frac{-1}{16}$ $\frac{\partial z}{\partial x}=f_{xy}(0,2)=0$ c. $f_{xx}(-1,0)=0$
We are given $z=f(x,y)=\frac{x+y^{2}}{x-y^{2}}$ $f_{x}(x,y) =\frac{(x-y^{2}).1-(x+y^{2}).1}{(x-y^{2})^{2}}=\frac{-2y^{2}}{(x-y^{2})^{2}}$ $f_{y}(x,y)=\frac{(x-y^{2}).2y-(x+y^{2}).(-2y)}{(x-y^{2})^{2}}=\frac{4xy}{(x-y^{2})^{2}}$ a. $\frac{\partial z}{\partial y}=f_{yy}(x,y)=\frac{(x-y^{2})^{2}.4x-4xy.2.(x-y^{2}).(-2y)}{(x-y^{2})^{4}}=\frac{4x^{3}-8x^{2}y^{2}+4xy^{4}+16x^{2}y^{2}-16xy^{4}}{(x-y^{2})^{4}}=\frac{4x^{3}+8x^{2}y^{2}-12xy^{4}}{(x-y^{2})^{4}}$ $\frac{\partial z}{\partial y}=f_{yx}(x,y)=\frac{(x-y^{2})^{2}.4y-4xy.2.(x-y^{2}).1}{(x-y^{2})^{4}}=\frac{4x^{2}y-8xy^{3}+4y^{5}-8x^{2}y+8xy^{3}}{(x-y^{2})^{4}}=\frac{4y^{5}-8x^{2}y+4x^{2}y}{(x-y^{2})^{4}}$ b. $\frac{\partial z}{\partial x}=f_{xx}(x,y)=\frac{(x-y^{2})^{2}.0-(-2y^{2}).(x-y^{2}).1}{(x-y^{2})^{4}}=\frac{2xy^{2}-2y^{3}}{(x-y^{2})^{4}}$ $\frac{\partial z}{\partial x}=f_{xy}(x,y)=\frac{(x-y^{2})^{2}.(-4y)-(-2y^{2}).(x-y^{2}).(-2y)}{(x-y^{2})^{4}}=\frac{-x^{2}y+8xy^{3}-4y^{5}-4xy^{3}+4y^{5}}{(x-y^{2})^{4}}=\frac{4xy^{3}-x^{2}y}{(x-y^{2})^{4}}$ $\frac{\partial z}{\partial x}=f_{xx}(0,2)=\frac{-1}{16}$ $\frac{\partial z}{\partial x}=f_{xy}(0,2)=0$ c. $\frac{\partial z}{\partial x}=f_{xx}(-1,0)=0$