Ex. 8: If \( u=\tan ^{-1}\left(\frac{x y}{\sqrt{1+x^{2}+y^{2}}}\right) \) then show that

Question

If \( u=\tan ^{-1}\left(\frac{x y}{\sqrt{1+x^{2}+y^{2}}}\right) \) then show that \( \frac{\partial^{2} u}{\partial x \partial y}=\frac{1}{\left(1+x^{2}+y^{2}\right)^{3 / 2}} \).

Answer 1

u = tan^-1(\(\frac{xy}{\sqrt{1+x^2+y^2}}\))

⇒ tan u = \(\frac{xy}{\sqrt{1+x^2+y^2}}\) 

Partial differentiate both sides w.r.t. y we get

sec²u \(\frac{\partial u}{\partial y} = \cfrac{x\sqrt{1+x^2+y^2}-xy\times\frac{2y}{2\sqrt{1+x^2+y^2}}}{1+x^2+y^2}\) 

⇒ sec²u \(\frac{\partial u}{\partial y} = \frac{x(1+x^2+y^2-y^2)}{(1+x^2+y^2)^{3/2}}\) 

 ⇒ sec²u \(\frac{\partial u}{\partial y} = \frac{x(1+x^2)}{(1+x^2+y^2)^{3/2}}\)

⇒   (1 + tan²u) \(\frac{\partial u}{\partial y} = \frac{x(1+x^2)}{(1+x^2+y^2)^{3/2}}\)

⇒ (1 + \(\frac{x^2y^2}{1+x^2+y^2}\)) \(\frac{\partial u}{\partial y}\)  = \( \frac{x(1+x^2)}{(1+x^2+y^2)^{3/2}}\)

⇒ (\(\frac{1+x^2+y^2+x^2y^2}{1+x^2+y^2}\))\(\frac{\partial u}{\partial y}\) = \( \frac{x(1+x^2)}{(1+x^2+y^2)^{3/2}}\)

⇒ \(\frac{\partial u}{\partial y}\) = \(\frac{x}{(1+x^2+y^2)^{1/2}(1+y^2)}\) 

By partial differentiating both sides w.r.t. x, we get

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

⇒ \(\frac{\partial u}{\partial x \partial y}=\frac1{1+y^2}\times\frac{1+x^2+y^2-x^2}{(1+x^2+y^2)^{3/2}}\) 

\(=\frac{1+y^2}{(1+y^2)(1+x^2+y^2)^{3/2}}\) = \(\frac{1}{(1+x^2+y^2)^{3/2}}\)

Hence Proved