Answer
$\frac{∂f}{∂x}=6y(2xy+1)^2$
$\frac{∂f}{∂y}=6x(2xy+1)^2$
$\frac{∂f}{∂x}|_{(1,-1)}=-6$
$\frac{∂f}{∂y}|_{(1,-1)}=6$
Work Step by Step
We find the partial derivatives as follows:
$f(x,y)=(2xy+1)^3$
$\frac{∂f}{∂x}=3(2xy+1)^2(2y)=6y(2xy+1)^2$
$\frac{∂f}{∂y}=3(2xy+1)^2(2x)=6x(2xy+1)^2$
$\frac{∂f}{∂x}|_{(1,-1)}=6(-1)(2(1)(-1)+1)^2=-6(-1)^2=-6$
$\frac{∂f}{∂y}|_{(1,-1)}=6(1)(2(1)(-1)+1)^2=6(-1)^2=6$
