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. $se{c}^{2}2u$
• D. $tan2u$

Answer 1

Answer: (A)

Euler's theorem $x\frac{\ell z}{\ell x}+y\frac{\ell z}{\ell y}=nz$
Given : U = tan${}^{-1}\frac{x^3+y^3}{x+y}$
$\Rightarrow tanU=\frac{x^3+y^3}{x+y}=z(let)$
$n=3-1=2$
$\therefore x\frac{\ell z}{\ell x}+y\frac{\ell z}{\ell y}=2z$
$\Rightarrow x\frac{\ell }{\ell x}tanU+y\frac{\ell }{\ell y}.tanU=2tanU$
$\Rightarrow x.se{c}^{2}U.\frac{\ell U}{\ell x}+y.se{c}^{2}U.\frac{\ell U}{\ell y}=2tanU$
$\Rightarrow se{c}^{2}U.[x\frac{\ell U}{\ell x}+y\frac{\ell U}{\ell y}]=2tanU$
$\Rightarrow x.\frac{\ell U}{\ell x}+y\frac{\ell U}{\ell y}=2.\frac{sinU}{cosU}.co{s}^{2}U$
$\Rightarrow x\frac{\ell U}{\ell x}+y\frac{\ell U}{\ell y}=sin2U$