Question

If u = tan$^{-1}${$\frac{x^3 + y^3}{x + y}$}, then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=$

• A. sin 2u
• B. cos 2u
• C. $sec^2 \,2u$
• D. $tan\, 2u$

Answer 1

Answer: (A)

Euler's theorem $x \frac{\ell z}{\ell x} + y \frac{\ell z}{\ell y}=nz$
Given : U = tan$^{-1} \, \, \frac{x^3 + y^3}{x+y}$
$\Rightarrow \, \, tan \, U \, = \, \frac{x^3 + y^3}{x+y} = z(let)$
$n=3-1=2$
$\therefore \, \, x\frac{\ell z}{\ell x} + y \frac{\ell z}{\ell y} = 2z$
$\Rightarrow \, \, x \frac{\ell}{\ell x} tan \, U \, + \, y \frac{\ell}{\ell y}.tan \, U \, = \, 2 tan U$
$\Rightarrow \, \, x.sec^2 \, U. \frac{\ell U}{\ell x} + y.sec^2 U . \frac{\ell U}{\ell y}=2 \, tan \, U$
$\Rightarrow \, \, \, sec^2 U. \bigg[ x \frac{\ell U}{\ell x} + y\frac{\ell U}{\ell y}\bigg] =2 tan U$
$\Rightarrow \, \, x. \frac{\ell U}{\ell x} + y \frac{\ell U}{\ell y} =2 . \frac{sin U}{cos U}. cos^2 U$
$\Rightarrow \, \, \, x \frac{\ell U}{\ell x} + y \frac{\ell U}{\ell y} \, = \, sin2 U$