Question

If \(u=x^2 tan^{-1} (\frac{y}{x})-y^2 tan^{-1} (\frac{x}{y})\) then \(\frac{∂^2 u}{∂x∂y}\) is?

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

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

(c) \(\frac{x^2}{x^2+y^2}\)

(d) \(\frac{y^2}{x^2+y^2}\)

I had been asked this question in an online interview.

Question is from Partial Differentiation in section Partial Differentiation of Engineering Mathematics

Answer 1

The correct answer is (b) \(\frac{x^2-y^2}{x^2+y^2}\)

The explanation is: \(\frac{∂^2 u}{∂x∂y}=\frac{∂}{∂x} \frac{∂u}{∂y}\)

\(\frac{∂u}{∂y}=-2y \,Tan^{-1} (\frac{x}{y})+\frac{x^3}{x^2+y^2}+\frac{xy^2}{x^2+y^2}\)

\(\frac{∂u}{∂y}=-2y \,Tan^{-1} (\frac{x}{y})+x\)

Now,

\(\frac{∂^2 u}{∂x∂y}=\frac{∂}{∂x} \frac{∂u}{∂y}= -\frac{2y}{1+\frac{x^2}{y^2}} (\frac{1}{y})+1=\frac{x^2-y^2}{x^2+y^2}\)