Question

Verify $\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{{\partial }^{2}u}{\partial y\partial x}$ for the function $u=\sin \left(\frac{x}{y}\right)$.

Answer 1

Given, $u=\sin \left(\frac{x}{y}\right)$
Therefore, $\frac{\partial u}{\partial x}=\frac{\cos \left(\frac{x}{y}\right)}{y}$
and $\frac{\partial u}{\partial y}=-x\frac{\cos \left(\frac{x}{y}\right)}{y^2}$
$=\frac{1}{y^2}[-x\times \sin \left(\frac{x}{y}\right)\left(\frac{1}{y}\right)-\cos \left(\frac{x}{y}\right)\times 1]$
$=\frac{x}{y^3}\sin \left(\frac{x}{y}\right)-\frac{1}{y^2}\cos \left(\frac{x}{y}\right)$ ........ (1)
$\frac{{\partial }^{2}u}{\partial y\partial x}=\frac{\partial }{\partial y}\left[\frac{\cos \left(\frac{x}{y}\right)}{y}\right]$
$=\frac{1}{y}[-\sin \left(\frac{x}{y}\right)\frac{-x}{y^2}]+\cos \left(\frac{x}{y}\right)(-\frac{1}{y^2})$
$\frac{{\partial }^{2}u}{\partial y\partial x}=\frac{x}{y^3}\sin \left(\frac{x}{y}\right)-\frac{1}{y^2}\cos \left(\frac{x}{y}\right)$ .......... (2)
From (1) & (2), we have

$\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{{\partial }^{2}u}{\partial y\partial x}$