AP EAMCET · Maths · Differentiation
If \(u=\sin ^{-1}\left(\frac{x}{y}\right)+\tan ^{-1}\left(\frac{y}{x}\right)\), then the value \(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}\) is
- A 0
- B 1
- C 2
- D None of these
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\begin{aligned} & \because u=\sin ^{-1}\left(\frac{x}{y}\right)+\tan ^{-1}\left(\frac{y}{x}\right) \\ & \therefore \frac{\partial u}{\partial x}=\frac{1}{\sqrt{\left\{1-\left(\frac{x}{y}\right)^2\right\}}} \frac{1}{y}+\frac{1}{1+\left(\frac{y}{x}\right)^2}…
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