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ExamsJEE MainMaths

If x and y are related by x²*y*(2x + y) = 3, find dy/dx.

  1. -(6x²*y + 2x*y²)/(2x³ + 2x²*y)
  2. -(2x³ + 2x²*y)/(6x²*y + 2x*y²)
  3. (6x²*y + 2x*y²)/(2x³ + 2x²*y)
  4. -(3x²*y + x*y²)/(x³ + x²*y)

Correct answer: -(6x²*y + 2x*y²)/(2x³ + 2x²*y)

Solution

Write the relation as F = 2x³*y + x²*y² = 3 (constant). Differentiating implicitly: d/dx(2x³*y) = 6x²*y + 2x³*y' and d/dx(x²*y²) = 2x*y² + 2x²*y*y'. Their sum is 0, giving (2x³ + 2x²*y)*y' = -(6x²*y + 2x*y²), so y' = -(6x²*y + 2x*y²)/(2x³ + 2x²*y).

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