If u = x2 tan-1 (x/y), show that ∂2u/∂x∂y = -x2(x2 + 3y2)/(x

1.	If u = x2 tan-1 (x/y), show that ∂2u/∂x∂y = -x2(x2 + 3y2)/(x2 + y2)2.
Answer» u = x² tan^-1\(\frac xy\) \(\frac{\partial u}{\partial y}=\cfrac{x^2}{1+\frac{x^2}{y^2}}\times\frac{-x}{y^2}\) = \(=\frac{-x^3}{x^2+y^2}\) \(\frac{\partial^2y}{\partial x\partial y}\) = \(-\frac{(x^2+y^2)(3x^2)-x^3(2x)}{(x^2+y^2)^2}\) = \(-\left(\frac{3x^4+3x^2y^2-2x^4}{(x^2+y^2)^2}\right)\) = \(-\frac{x^2(x^2+3y^2)}{(x^2+y^2)^2}\) \(\frac{\partial u}{\partial y}=\cfrac{x^2}{1+\frac{x^2}{y^2}}\times\frac 1y\) + 2x tan^-1\(\frac xy\) = \(\frac{x^2y}{x^2+y^2}+2xtan^{-1}\frac xy\) \(\frac{\partial^2y}{\partial x\partial y}\) = \(\frac{(x^2+y^2)x^2-x^2y(2y)}{(x^2+y^2)^2}\) + \(\cfrac{2x}{1+\frac{x^2}{y^2}}\times\frac{-x}{y^2}\) = \(\frac{x^4+x^2y^2-2x^2y^2}{(x^2+y^2)^2}-\frac{2x^2}{x^2+y^2}\) = \(\frac{x^4-x^2y^2-2x^2(x^2+y^2)}{(x^2+y^2)^2}\) = \(\frac{-x^4-3x^2y^2}{(x^2+y^2)^2}=\frac{-x^2(x^2+3y^2)}{(x^2+y^2)^2}\) \(\therefore\) \(\frac{\partial^2 y}{\partial x\partial y} = \frac{\partial y}{\partial y\partial x}\) Hence Proved.

If u = x2 tan-1 (x/y), show that ∂2u/∂x∂y = -x2(x2 + 3y2)/(x2 + y2)2.

Answer»

u = x² tan^-1\(\frac xy\)

\(\frac{\partial u}{\partial y}=\cfrac{x^2}{1+\frac{x^2}{y^2}}\times\frac{-x}{y^2}\) = \(=\frac{-x^3}{x^2+y^2}\)

\(\frac{\partial^2y}{\partial x\partial y}\) = \(-\frac{(x^2+y^2)(3x^2)-x^3(2x)}{(x^2+y^2)^2}\)

= \(-\left(\frac{3x^4+3x^2y^2-2x^4}{(x^2+y^2)^2}\right)\) = \(-\frac{x^2(x^2+3y^2)}{(x^2+y^2)^2}\)

\(\frac{\partial u}{\partial y}=\cfrac{x^2}{1+\frac{x^2}{y^2}}\times\frac 1y\) + 2x tan^-1\(\frac xy\) = \(\frac{x^2y}{x^2+y^2}+2xtan^{-1}\frac xy\)

\(\frac{\partial^2y}{\partial x\partial y}\) = \(\frac{(x^2+y^2)x^2-x^2y(2y)}{(x^2+y^2)^2}\) + \(\cfrac{2x}{1+\frac{x^2}{y^2}}\times\frac{-x}{y^2}\)

= \(\frac{x^4+x^2y^2-2x^2y^2}{(x^2+y^2)^2}-\frac{2x^2}{x^2+y^2}\)

= \(\frac{x^4-x^2y^2-2x^2(x^2+y^2)}{(x^2+y^2)^2}\)

= \(\frac{-x^4-3x^2y^2}{(x^2+y^2)^2}=\frac{-x^2(x^2+3y^2)}{(x^2+y^2)^2}\)

\(\therefore\) \(\frac{\partial^2 y}{\partial x\partial y} = \frac{\partial y}{\partial y\partial x}\) Hence Proved.