answered
2014-07-08 12:01:49 -0600
Searching through the textbook, I found this piece of information on p. 371 (14.6):
“mixed partials” [are] computed by taking partial derivatives with respect to both variables in the two possible orders
They are notated $f_{xy}$ and $f_{yx}$.
To find $f_{xy}$, we take the $x$ partial derivative then the $y$ partial derivative of $f$:
$$ f_x = 3x^2y^4 $$
$$ f_{xy} = (3x^2)(4y^3) = 12x^2y^3$$
Note that we leave the $x$'s alone in the second step and only focus on the $y$'s. We will use a very similar process to find $f_x{yx}$:
$$ f_y = 4y^3x^3 $$
$$ f_{yx} = (4y^3)(3x^2) = 12x^2y^3 $$
$$ f_{xy} = f_{yx} = 12x^2y^3 $$
Theorem 14.6.2 (Clairaut's Theorem) then says:
If the mixed partial derivatives are continuous, they are equal.
Which is a more generalized version of what we just showed.