Answer
a) $\frac{\partial f(x,y)}{\partial x}=3\times2\times x-0+1-0=6x+1$
b) $\frac{\partial f(x,y)}{\partial y}=0-3\times y^2+0-0=-3y^2$
c) $\frac{\partial f(x,y)}{\partial x} (1;-1)=3\times2\times x-0+1-0=6x+1=6\times(1)+1=8$
d) $\frac{\partial f(x,y)}{\partial y} (1;-1)=0-3\times y^2+0-0=-3y^2=-3\times(-1)^2=-3$
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:
$\frac{\partial f(x,y)}{\partial x}=3\times2\times x-0+1-0=6x+1$
b) Based on the previous exercise.
$\frac{\partial f(x,y)}{\partial y}=0-3\times y^2+0-0=-3y^2$
c) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial x} (1;-1)=3\times2\times x-0+1-0=6x+1=6\times(1)+1=8$
d) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial y} (1;-1)=0-3\times y^2+0-0=-3y^2=-3\times(-1)^2=-3$
