Answer to Question #129749 in Calculus for gazal

Question

If u = tan inverse ((x cube - y cube) /(x cube + y cube)) .Then show that x.ux + y.uy = 0

Answer 1

As per the question,

u=$tan^{-1}\frac{x^3-y^3}{x^3+y^3}$

differentiate u with respect to x

$\frac{du}{dx}=\frac{1}{1+(\frac{x^3-y^3}{x^3+y^3})^2}\times\frac{(x^3+y^3)3x^2-(x^3-y^3)3x^2}{(x^3+y^3)^2}$

on solving above expression we get,

$u_x=\frac{1}{(x^3+y^3)^2+(x^3-y^3)^2}\times(3x^2y^3+3x^2y^3)$

$u_x=\frac{1}{(x^3+y^3)^2+(x^3-y^3)^2}\times(6x^2y^3)$ ....(1)

Now differentiating u with respect to y.

$\frac{du}{dy}=\frac{1}{1+(\frac{x^3-y^3}{x^3+y^3})^2}\times\frac{(x^3+y^3)(-3y^2)-(x^3-y^3)3y^2}{(x^3+y^3)^2}$

on solving we get,

$u_y=\frac{1}{(x^3+y^3)^2+(x^3-y^3)^2}\times(-3x^3y^2-3x^3y^2)$

$u_y=\frac{1}{(x^3+y^3)^2+(x^3-y^3)^2}\times(-6x^3y^2)$ ...(2)

To show $xu_x+yu_y=0$

$\frac{x}{(x^3+y^3)^2+(x^3-y^3)^2}\times(6x^2y^3)$ $+\frac{y}{(x^3+y^3)^2+(x^3-y^3)^2}\times(-6x^3y^2)$

=$\frac{1}{(x^3+y^3)^2+(x^3-y^3)^2}\times(6x^3y^3-6x^3y^3)$

=0

Hence $xu_x+yu_y=0$ .