Step 1
Mixed Derivative theorem:" If the function f(x,y) and its partial derivatives \(\displaystyle{f}_{{x}},{f}_{{y}},{f}_{{{x}{y}}}{\quad\text{and}\quad}{f}_{{{y}{x}}}\) are all defined in any open interval (a,b) and all are continues in the interval, then \(\displaystyle{{f}_{{{x}{y}}}{\left({a},{b}\right)}}={{f}_{{{y}{x}}}{\left({a},{b}\right)}}\)".
That is, mixed derivative theorem says that the mixed partial derivatives are equal.
Thus, there is no need of calculating all the mixed partial derivatives. Only one case is enough.
Step 2
For example consider the function \(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\).
Find the first order partial derivatives as follows.
\(\displaystyle\frac{\partial}{{\partial{x}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{x}}}{\left({x}^{{3}}{y}^{{3}}\right)}+\frac{\partial}{{\partial{x}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{x}}}{\left({y}^{{2}}{x}\right)}\)
\(\displaystyle={3}{y}^{{3}}{x}^{{2}}+{2}{y}{x}+{y}^{{2}}\)
\(\displaystyle\frac{\partial}{{\partial{y}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{y}}}{\left({x}^{{3}}{y}^{{3}}\right)}+\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{y}}}{\left({y}^{{2}}{x}\right)}\)
\(\displaystyle={3}{x}^{{3}}{y}^{{2}}+{x}^{{2}}+{2}{x}{y}\)
Step 3
Now, find the mixed partial derivatives as,
\(\displaystyle\frac{\partial^{{2}}}{{\partial{y}\partial{x}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{x}}}{\left({3}{x}^{{3}}{y}^{{2}}+{x}^{{2}}+{2}{x}{y}\right)}\)
\(\displaystyle={9}{y}^{{2}}{x}^{{2}}+{2}{x}+{2}{y}\)
\(\displaystyle\frac{\partial^{{2}}}{{\partial{x}\partial{y}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{y}}}{\left({3}{y}^{{3}}{x}^{{2}}+{2}{y}{x}+{y}^{{2}}\right)}\)
\(\displaystyle={9}{x}^{{2}}{y}^{{2}}+{2}{x}+{2}{y}\)
That is, \(\displaystyle\frac{{\partial^{{2}}{f}}}{{\partial{y}\partial{x}}}=\frac{{\partial^{{2}}{f}}}{{\partial{x}\partial{y}}}\).
Mixed Derivative theorem:" If the function f(x,y) and its partial derivatives \(\displaystyle{f}_{{x}},{f}_{{y}},{f}_{{{x}{y}}}{\quad\text{and}\quad}{f}_{{{y}{x}}}\) are all defined in any open interval (a,b) and all are continues in the interval, then \(\displaystyle{{f}_{{{x}{y}}}{\left({a},{b}\right)}}={{f}_{{{y}{x}}}{\left({a},{b}\right)}}\)".
That is, mixed derivative theorem says that the mixed partial derivatives are equal.
Thus, there is no need of calculating all the mixed partial derivatives. Only one case is enough.
Step 2
For example consider the function \(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\).
Find the first order partial derivatives as follows.
\(\displaystyle\frac{\partial}{{\partial{x}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{x}}}{\left({x}^{{3}}{y}^{{3}}\right)}+\frac{\partial}{{\partial{x}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{x}}}{\left({y}^{{2}}{x}\right)}\)
\(\displaystyle={3}{y}^{{3}}{x}^{{2}}+{2}{y}{x}+{y}^{{2}}\)
\(\displaystyle\frac{\partial}{{\partial{y}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{y}}}{\left({x}^{{3}}{y}^{{3}}\right)}+\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{y}}}{\left({y}^{{2}}{x}\right)}\)
\(\displaystyle={3}{x}^{{3}}{y}^{{2}}+{x}^{{2}}+{2}{x}{y}\)
Step 3
Now, find the mixed partial derivatives as,
\(\displaystyle\frac{\partial^{{2}}}{{\partial{y}\partial{x}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{x}}}{\left({3}{x}^{{3}}{y}^{{2}}+{x}^{{2}}+{2}{x}{y}\right)}\)
\(\displaystyle={9}{y}^{{2}}{x}^{{2}}+{2}{x}+{2}{y}\)
\(\displaystyle\frac{\partial^{{2}}}{{\partial{x}\partial{y}}}{\left({x}^{{3}}{y}^{{3}}+{x}^{{2}}{y}+{y}^{{2}}{x}\right)}=\frac{\partial}{{\partial{y}}}{\left({3}{y}^{{3}}{x}^{{2}}+{2}{y}{x}+{y}^{{2}}\right)}\)
\(\displaystyle={9}{x}^{{2}}{y}^{{2}}+{2}{x}+{2}{y}\)
That is, \(\displaystyle\frac{{\partial^{{2}}{f}}}{{\partial{y}\partial{x}}}=\frac{{\partial^{{2}}{f}}}{{\partial{x}\partial{y}}}\).
