Answer
$\dfrac{∂f}{∂x}=yz$
$\dfrac{∂f}{∂y}=xz$
and $\dfrac{∂f}{∂z}=xy$
$\dfrac{∂f}{∂x}|_{(0,-1,1)}=-1$ and $\dfrac{∂f}{∂y}|_{(0,-1,1)}=0$
and $\dfrac{∂f}{∂z}|_{(0,-1,1)}=0$
Work Step by Step
Given: $f(x,y,z)=xyz$
We will find the partial derivatives as follows:
$\dfrac{∂f}{∂x}=yz\dfrac{∂f}{∂x}(x)=yz$
$\dfrac{∂f}{∂y}=xz\dfrac{∂f}{∂y}(y)=xz$
and $\dfrac{∂f}{∂z}=xy\dfrac{∂f}{∂x}(z)=xy$
$\dfrac{∂f}{∂x}|_{(0,-1,1)}=(-1)(1)=-1$ and $\dfrac{∂f}{∂y}|_{(0,-1,1)}=(0)(1)=0$
and $\dfrac{∂f}{∂z}|_{(0,-1,1)}=(0)(1)=0$
