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CUET · MATHS · PYQ PAPER 2025

If \(x y+\frac{x^2}{y}=x^3 y+y\), then \(\frac{d y}{d x}\) is equal to

  1. A \(\frac{y\left(3 x^2 y^2+2 x+y^2\right)}{x^2 y-x^2 y^3+x^2-y^2}\)
  2. B \(\frac{y\left(3 x^2 y^2-2 x-y^2\right)}{x y^2-x^3 y^2-x^2-y^2}\)
  3. C \(\frac{y\left(3 x^2 y^2-2 x+y^2\right)}{x^2 y-x^2 y^3+x^2+y^2}\)
  4. D \(\frac{y\left(3 y^2 x^2-y^2+2 x\right)}{x^2 y^3-x^2 y-x-y^2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{y\left(3 x^2 y^2-2 x-y^2\right)}{x y^2-x^3 y^2-x^2-y^2}\)

Step-by-step Solution

Detailed explanation

\(x y+x^2 y^{-1}=x^3 y+y\) \(\frac{d}{dx}(xy) + \frac{d}{dx}(x^2 y^{-1}) = \frac{d}{dx}(x^3 y) + \frac{d}{dx}(y)\) \(y + x \frac{dy}{dx} + \frac{2x}{y} - \frac{x^2}{y^2} \frac{dy}{dx} = 3x^2 y + x^3 \frac{dy}{dx} + \frac{dy}{dx}\)…
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