Q64 / 120 · VITEEE 2006 · Mathematics
If $u=\tan ^{-1}\left\{\frac{x^{3}+y^{3}}{x+y}\right\}$, then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=$
- Correct answerA$\sin 2 \mathrm{u}$
- B$\cos 2 \mathrm{u}$
- C$\sec ^{2} 2 \mathrm{u}$
- D$\tan 2 \mathrm{u}$
Solution
Euler's theorem $x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=n z$
$\begin{array}{l}
\text { Given : } U=\tan ^{-1} \frac{x^{3}+y^{3}}{x+y} \\
\Rightarrow \tan U=\frac{x^{3}+y^{3}}{x+y}=z \text { (let) } \\
n=3-1=2 \\
\therefore x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=2 z
\end{array}$
$\Rightarrow \mathrm{x} \frac{\partial}{\partial \mathrm{x}} \tan \mathrm{U}+\mathrm{y} \frac{\partial}{\partial \mathrm{y}} \cdot \tan \mathrm{U}=2 \tan \mathrm{U}$
$\Rightarrow \mathrm{x} \cdot \sec ^{2} \mathrm{U} \cdot \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \cdot \sec ^{2} \mathrm{U} \cdot \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=2 \tan \mathrm{U}$
$\Rightarrow \sec ^{2} \mathrm{U} \cdot\left[\mathrm{x} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}\right]=2 \tan \mathrm{U}$
$\Rightarrow \mathrm{x} \cdot \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=2 \cdot \frac{\sin \mathrm{U}}{\cos \mathrm{U}} \cdot \cos ^{2} \mathrm{U}$
$\Rightarrow \mathrm{x} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\sin 2 \mathrm{U}$
Hence, the correct answer is option: 'A'.