AP EAMCET · Maths · Differentiation
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 \(\cot z\)
- B \(2 \cot z\)
- C \(2 \tan z\)
- D \(2 \sec z\)
Answer & Solution
Correct Answer
(B) \(2 \cot z\)
Step-by-step Solution
Detailed explanation
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, \(\quad n=2\)…
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