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RD Sharma - Volume 1

11. Differentiation

Exercise 11.1

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Exercise 11.2

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Exercise 11.3

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Exercise 11.4

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Exercise 11.5

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Exercise 11.6

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Exercise 11.7

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Exercise 11.8

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MCQ

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Very short answer

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Q. 2

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Class 12th RD Sharma - Volume 1 11. Differentiation

Answer :

We are given with an equation y³ – 3xy² = x³ + 3x²y, we have to find of it, so by differentiating the equation on both sides with respect to x, we get,

3y² ‐ 3[y²(1) + 2xy ] = 3x² + 3[2xy + x²]

Taking terms to left hand side and taking common , we get,

[3y² ‐ 6xy ‐ 3x²] = 3x² + 6xy + 3y²

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