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If x and y are related by x²*y*(2x + y) = 3, find dy/dx.
- -(6x²*y + 2x*y²)/(2x³ + 2x²*y)
- -(2x³ + 2x²*y)/(6x²*y + 2x*y²)
- (6x²*y + 2x*y²)/(2x³ + 2x²*y)
- -(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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