NTA Abhyas JEE Main2020MathematicsDifferentiationPractice

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

Options

A- 2 1 + x 2 for all x
B2 1 + x 2 for all x < 1
C2 1 + x 2 for x > 1
DNone of these

Correct answer

B. 2 1 + x 2 for all x < 1

Step-by-step solution

d y d x = 1 1 - 4 x 2 1 + x 2 2 d d x 2 x 1 + x 2 ⇒ d y d x = 1 + x 2 1 - x 2 2 2 1 + x 2 - 4 x 2 1 + x 2 2 ⇒ d y d x = 2 1 + x 2 1 - x 2 1 - x 2 1 + x 2 2 ⇒ d y d x = 2 1 - x 2 1 - x 2 1 1 + x 2 ⇒ d y d x = - 2 1 + x 2 , i f x > 1 2 1 + x 2 , i f x < 1

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