Answer
a) $\frac{\partial f(x,y)}{\partial x}=0.5\times x^{-0.5}-0+0+0=0.5\times x^{-0.5}$
b) $\frac{\partial f(x,y)}{\partial y}=0-2\times4\times y^3+1+0=-8y^3+1$
c) $\frac{\partial f(x,y)}{\partial x} (1;-1)=0.5\times x^{-0.5}-0+0+0=0.5\times x^{-0.5}=0.5\times(1)^{-0.5}=0.5$
d) $\frac{\partial f(x,y)}{\partial y} (1;-1)=0-2\times4\times y^3+1+0=-8y^3+1= -8\times(-1)^3+1=9$
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:
$\frac{\partial f(x,y)}{\partial x}=0.5\times x^{-0.5}-0+0+0=0.5\times x^{-0.5}$
b) Based on the previous exercise.
$\frac{\partial f(x,y)}{\partial y}=0-2\times4\times y^3+1+0=-8y^3+1$
c) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial x} (1;-1)=0.5\times x^{-0.5}-0+0+0=0.5\times x^{-0.5}=0.5\times(1)^{-0.5}=0.5$
d) Now, we have to substitute $(1; -1)$ in to the variables $x$ and $y$ in the partial derivative, respectively.
$\frac{\partial f(x,y)}{\partial y} (1;-1)=0-2\times4\times y^3+1+0=-8y^3+1= -8\times(-1)^3+1=9$
