Question

If $u=\sin xy$,

Find:$\frac{{\partial }^{2}u}{{\partial }_y{\partial }_x}$

Answer 1

$u=\sin \left(xy\right)$

$\frac{\partial u}{\partial x}=\frac{\partial \left[\sin \right(xy\left)\right]}{\partial x}$

$=\cos \left(xy\right)y$

$\frac{{\partial }^{2}u}{\partial y\partial x}=\frac{{\partial }^2\left[\cos \right(xy\left)y\right]}{\partial y\partial x}$

$=-xy\sin \left(xy\right)+\cos \left(xy\right)$

$\frac{{\partial }^{2}u}{\partial y\partial x}=-xy\left(u\right)+\frac{1}{y}\frac{\partial u}{\partial x}$