Q110 / 160 · TS EAMCET 2008 · Mathematics
If $\quad z=\sec ^{-1}\left(\frac{x^4+y^4-8 x^2 y^2}{x^2+y^2}\right), \quad$ then $x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}$ is equal to
- A$\cot z$
- Correct answerB$2 \cot z$
- C$2 \tan z$
- D$2 \sec z$
Solution
Given that,
$$
\begin{aligned}
z & =\sec ^{-1}\left(\frac{x^4+y^4-8 x^2 y^2}{x^2+y^2}\right) \\
\Rightarrow \quad \sec z & =\frac{x^4+y^4-8 x^2 y^2}{x^2+y^2}
\end{aligned}
$$
Here, $\quad n=2$
$$
\begin{aligned}
\therefore \quad x \frac{\delta z}{\delta x}+y \frac{\delta z}{\delta y} & =2 \cdot \frac{\sec z}{\sec z \cdot \tan z} \\
& =2 \cot z
\end{aligned}
$$
Hence, the correct answer is option: 'B'.