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