Question

14. Consider the function f(x, y) = yx^4 * e^x². Find fxyxyx (x, y).​

Answer 1

The second partial derivatives of $f$ exist everywhere in its domain. By Schwarz's theorem, the mixed second-order partials derivatives are equal:

$f_{xy} = f_{yx}$

Then

$f(x,y) = y x^4 e^{x^2} \\\\ \implies f_y = x^4 e^{x^2} \\\\ \implies f_{yx} = f_{xy} = g(x) \\\\ \implies f_{xyx} = g'(x) \\\\ \implies f_{xyxy} = 0 \\\\ \implies f_{xyxyx} = \boxed{0}$

We don't actually need to compute $g$ and $g'$ to know that they are both free of $y$. So when we differentiate for a second time with respect to $y$, the whole thing cancels out.