## Calculus 8th Edition

$$f_x = 2xy - 0 = 2xy$$ $$f_y = x^2 - 12xy$$
To find $f_x$, we treat $y$ as a constant and differentiate with respect to $x$. Thus $$f_x = 2xy - 0 = 2xy.$$ Similarly, to find $f_y$, we treat $x$ s a constant and differentiate with respect to $y$. Thus $$f_y = x^2 - 12xy.$$