Answer
$\frac{∂f}{∂x}=0.2ye^{0.2xy}$
$\frac{∂f}{∂y}=0.2xe^{0.2xy}$
$\frac{∂f}{∂x}|_{(1,-1)}=-0.2e^{-0.2}$
$\frac{∂f}{∂y}|_{(1,-1)}=0.2e^{-0.2}$
Work Step by Step
We find the partial derivatives as follows:
$f(x,y)=e^{0.2xy}$
$\frac{∂f}{∂x}=e^{0.2xy}(0.2y)=0.2ye^{0.2xy}$
$\frac{∂f}{∂y}=e^{0.2xy}(0.2x)=0.2xe^{0.2xy}$
$\frac{∂f}{∂x}|_{(1,-1)}=0.2(-1)e^{0.2(1)(-1)}=-0.2e^{-0.2}$
$\frac{∂f}{∂y}|_{(1,-1)}=0.2(1)e^{0.2(1)(-1)}=0.2e^{-0.2}$
