CUET · MATHS · PYQ PAPER 2023
If \(x\sqrt{1+y} + y\sqrt{1+x} = 0\), then \(\frac{dy}{dx}\) is:
- A \(\frac{1}{(1+x)^2}\)
- B \(-\frac{1}{(1+x)^2}\)
- C \((1+x)^2\)
- D \(-(1+x)^2\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{(1+x)^2}\)
Step-by-step Solution
Detailed explanation
\(x\sqrt{1+y} = -y\sqrt{1+x}\) \(x^2(1+y) = y^2(1+x)\) \(x^2+x^2y = y^2+y^2x\) \(x^2-y^2 = y^2x-x^2y\) \((x-y)(x+y) = -xy(x-y)\) \(x+y = -xy\) \(y(1+x) = -x\) \(y = -\frac{x}{1+x}\) \(\frac{dy}{dx} = -\frac{(1)(1+x) - x(1)}{(1+x)^2}\) \(\frac{dy}{dx} = -\frac{1+x-x}{(1+x)^2}\)…
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