Answer
a) $\frac{\partial f(x,y)}{\partial x}=4\times x^3\times y^2-1=4x^3y^2-1$
b) $\frac{\partial f(x,y)}{\partial y}=x^4\times2\times y-0=2x^4y$
c) $\frac{\partial f(x,y)}{\partial x} (1;-1)=4\times x^3\times y^2-1=4x^3y^2-1=4\times(1)^3\times(-1)^2=4$
d) $\frac{\partial f(x,y)}{\partial y} (1;-1)=x^4\times2\times y-0=2x^4y=2\times (1)^4\times(-1)=-2$
Work Step by Step
a)
$\frac{\partial f(x,y)}{\partial x}=4\times x^3\times y^2-1=4x^3y^2-1$
b) Based on the previous exercise.
$\frac{\partial f(x,y)}{\partial y}=x^4\times2\times y-0=2x^4y$
c) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial x} (1;-1)=4\times x^3\times y^2-1=4x^3y^2-1=4\times(1)^3\times(-1)^2=4$
d) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial y} (1;-1)=x^4\times2\times y-0=2x^4y=2\times (1)^4\times(-1)=-2$
