TS EAMCET · Maths · Differentiation
If \(u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)\), then \(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}\) is equal to
- A \(3 u\)
- B \(4 u\)
- C \(3 \sin u\)
- D \(3 \tan u\)
Answer & Solution
Correct Answer
(D) \(3 \tan u\)
Step-by-step Solution
Detailed explanation
\(u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)\) Let \(v=\sin u=\frac{x^4+y^4}{x+y}\), here degree is homogeneous, so \(n=4-1=3\) By Euler's theorem, we have to prove that, \(x \frac{\partial v}{\partial x}+y \frac{\partial v}{\partial y}=3 v\)…
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