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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Differentiation

JEE Main 2019 · 12 April · Evening Shift · Mathematics · Differentiation

The derivative of ^ - 1 (( x - x)/( x + x)) with respect to...

The derivative of tan⁡−1(sin⁡x−cos⁡xsin⁡x+cos⁡x)\tan^{- 1}\left(\dfrac{\sin x - \cos x}{\sin x + \cos x}\right) with respect to x2,\dfrac{x}{2}, where x∈(0,π2)x \in \left(0, \dfrac{\pi}{2}\right), is

Options

  1. A22
  2. B12\dfrac{1}{2}
  3. C23\dfrac{2}{3}
  4. D11

Correct answer

A. 22

Step-by-step solution

Given function can be written as

y=tan⁡−1(sin⁡x−cos⁡xsin⁡x+cos⁡x)y = \tan^{- 1}\left(\dfrac{\sin x - \cos x}{\sin x + \cos x}\right)

⇒y=tan⁡−1(tan⁡x−1tan⁡x+1)⇒ y = \tan^{- 1}\left(\dfrac{\tan x - 1}{\tan x + 1}\right)

⇒y=tan⁡−1(tan⁡(x−π4))(givenx∈(0,π2))⇒ y = \tan^{- 1}\left(\tan \left(x - \dfrac{\pi}{4}\right)\right)\left(givenx \in \left(0, \dfrac{\pi}{2}\right)\right)
⇒y=x−π4⇒ y = x - \dfrac{\pi}{4}
Now dyd(x2)=2dydx=2\dfrac{dy}{d\left(\dfrac{x}{2}\right)} = 2\dfrac{dy}{dx} = 2

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