Answer
a) $\frac{\partial f(x,y)}{\partial x}=-1\times x^{-2}\times y^2+y^2+y=-x^{-2}y^2+y^2+y$
b) $\frac{\partial f(x,y)}{\partial y}=x^{-1}\times2\times y+x\times2\times y+x=2x^{-1}y+2xy+x$
c) $\frac{\partial f(x,y)}{\partial x} (1;-1)=-1\times x^{-2}\times y^2+y^2+y=-x^{-2}y^2+y^2+y=-(1)^{-2}(-1)^2+(-1)^2+(-1)=-1$
d) $\frac{\partial f(x,y)}{\partial y} (1;-1)=x^{-1}\times2\times y+x\times2\times y+x=2x^{-1}y+2xy+x=2(1)^{-1}(-1)+2(1)(-1)+1=-3$
Work Step by Step
a)
$\frac{\partial f(x,y)}{\partial x}=-1\times x^{-2}\times y^2+y^2+y=-x^{-2}y^2+y^2+y$
b) Based on the previous exercise.
$\frac{\partial f(x,y)}{\partial y}=x^{-1}\times2\times y+x\times2\times y+x=2x^{-1}y+2xy+x$
c) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial x} (1;-1)=-1\times x^{-2}\times y^2+y^2+y=-x^{-2}y^2+y^2+y=-(1)^{-2}(-1)^2+(-1)^2+(-1)=-1$
d) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial y} (1;-1)=x^{-1}\times2\times y+x\times2\times y+x=2x^{-1}y+2xy+x=2(1)^{-1}(-1)+2(1)(-1)+1=-3$
