TS EAMCET · Maths · Differentiation
If \(u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)\) then \(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}\) equal to :
- A \(\sin u\)
- B \(\tan u\)
- C \(\cos u\)
- D \(\cot u\)
Answer & Solution
Correct Answer
(B) \(\tan u\)
Step-by-step Solution
Detailed explanation
\(u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)\) Here \(u\) is not a homogeneous function. But \(f(x, y)=\sin u=\frac{x^2+y^2}{x+y}\) is a homogeneous function of degree one. Here by Euler's theorem \(x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=f\)…
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