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