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NTA Abhyas JEE Main2020MathematicsDifferentiationPractice

If y = 1 + x y + s i n - 1 s i n 2 x , then d y d x at x = 0 is

Options

  1. A0
  2. Bln ⁡ 2
  3. C1
  4. D1 2

Correct answer

C. 1

Step-by-step solution

d y d x = 1 + x y y 1 + x + ln ⁡ 1 + x ⋅ d y d x + s i n 2 x 1 - s i n 4 x Putting x = 0 & y = 1 , we get, d y d x = 1 1 + ln ⁡ 1 . d y d x + 0 = 1

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