Answer
a) $\frac{\partial f(x,y)}{\partial x}=0-40+0+10y=-40+10y$
b) $\frac{\partial f(x,y)}{\partial y}=0-0+20+10x=20+10x$
c) $\frac{\partial f(x,y)}{\partial x} (1;-1)=0-40+0+10y=-40+10y=-40+10\times(-1)=-50$
d) $\frac{\partial f(x,y)}{\partial y} (1;-1)=0-0+20+10x=20+10x=20+10\times(1)=30$
Work Step by Step
a) The partial derivative of $f(x,y)$ with respect to $x$ is calculated by derivating the the function as if $y$ was a constant.
That gives us:
$\frac{\partial f(x,y)}{\partial x}=0-40+0+10y=-40+10y$
b) Based on the previous exercise.
$\frac{\partial f(x,y)}{\partial y}=0-0+20+10x=20+10x$
c) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial x} (1;-1)=0-40+0+10y=-40+10y=-40+10\times(-1)=-50$
d) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial y} (1;-1)=0-0+20+10x=20+10x=20+10\times(1)=30$
