Answer
a) $\frac{\partial f(x,y)}{\partial x}=2\times x\times y^3-3\times x^2\times y^2-y=2xy^3-3x^2y^2-y$
b) $\frac{\partial f(x,y)}{\partial y}=x^2\times3\times y^2-x^3\times2\times y-x=3x^2y^2-2x^3y-x$
c) $\frac{\partial f(x,y)}{\partial x} (1;-1)=2\times x\times y^3-3\times x^2\times y^2-y=2xy^3-3x^2y^2-y=2\times(1)\times(-1)^3-3\times(1)^2\times(-1)^2-(-1)=-4$
d) $\frac{\partial f(x,y)}{\partial y} (1;-1)=x^2\times3\times y^2-x^3\times2\times y-x=3x^2y^2-2x^3y-x=3(1)^2(-1)^2-2(1)^3(-1)-1=4$
Work Step by Step
a)
$\frac{\partial f(x,y)}{\partial x}=2\times x\times y^3-3\times x^2\times y^2-y=2xy^3-3x^2y^2-y$
b) Based on the previous exercise.
$\frac{\partial f(x,y)}{\partial y}=x^2\times3\times y^2-x^3\times2\times y-x=3x^2y^2-2x^3y-x$
c) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial x} (1;-1)=2\times x\times y^3-3\times x^2\times y^2-y=2xy^3-3x^2y^2-y=2\times(1)\times(-1)^3-3\times(1)^2\times(-1)^2-(-1)=-4$
d) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial y} (1;-1)=x^2\times3\times y^2-x^3\times2\times y-x=3x^2y^2-2x^3y-x=3(1)^2(-1)^2-2(1)^3(-1)-1=4$
