Question:
If \(x\sqrt{1+y}+y\sqrt{1+x}=0;x\ne y\), then the value of \(\frac{d^2y}{dx^2}+2\frac{dy}{dx}\) at x = 1 is :
If \(x\sqrt{1+y}+y\sqrt{1+x}=0;x\ne y\), then the value of \(\frac{d^2y}{dx^2}+2\frac{dy}{dx}\) at x = 1 is :
Updated On: Jun 13, 2024
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- \(\frac{1}{4}\)
- \(-\frac{1}{4}\)
- \(\frac{1}{8}\)
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The Correct Option is C
Solution and Explanation
The correct option is (C) :\(-\frac{1}{4}\).
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