# Compute all of the second-order partial derivatives for the functions and show that the mixed partial derivatives are equal. f(x,y)=e^{x}\sin(xy)

Derivatives
Compute all of the second-order partial derivatives for the functions and show that the mixed partial derivatives are equal.
$$\displaystyle{f{{\left({x},{y}\right)}}}={e}^{{{x}}}{\sin{{\left({x}{y}\right)}}}$$

2021-05-14
Step 1
Given function is:
$$\displaystyle{f{{\left({x},{y}\right)}}}={e}^{{{x}}}{\sin{{\left({x}{y}\right)}}}$$
Differentiating f(x, y) partially with respect to x we get,
$$\displaystyle{{f}_{{{x}}}{\left({x},{y}\right)}}={e}^{{{x}}}{\sin{{\left({x}{y}\right)}}}+{y}{e}^{{{x}}}{\cos{{\left({x}{y}\right)}}}={e}^{{{x}}}{\left[{\sin{{\left({x}{y}\right)}}}+{y}{\cos{{\left({x}{y}\right)}}}\right]}$$
Differentiating f(x, y) partially with respect to y we get,
$$\displaystyle{{f}_{{{y}}}{\left({x},{y}\right)}}={x}{e}^{{{x}}}{\cos{{\left({x}{y}\right)}}}$$
Step 2
Moreover we get,
$$\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}={x}{e}^{{{x}}}{\cos{{\left({x}{y}\right)}}}+{e}^{{{x}}}{\cos{{\left({x}{y}\right)}}}-{x}{y}{e}^{{{x}}}{\sin{{\left({x}{y}\right)}}}$$
$$\displaystyle{{f}_{{{y}{x}}}{\left({x},{y}\right)}}={e}^{{{x}}}{\cos{{\left({x}{y}\right)}}}+{x}{e}^{{{x}}}{\cos{{\left({x}{y}\right)}}}-{x}{y}{e}^{{{x}}}{\sin{{\left({x}{y}\right)}}}$$
These are the second order partial derivatives of f(x, y).