Verified Solution • 2025–26 Session moderate

QJEEM19Ma09a2: JEE Main 2019 Mathematics - Class 12 JEEM19Ma09a2 Solution: The derivative of , with respect to , where is :

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

Multiple Choice Options

  • A 1
  • B 2
  • C 23{2 \over 3}
  • D 12{1 \over 2}

Step-by-Step Verified Solution

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

⇒\Rightarrow −tan⁡−1(1−tan⁡x1+tan⁡x)- {\tan ^{ - 1}}\left( {{{1 - \tan x} \over {1 + \tan x}}} \right)

⇒\Rightarrow −tan⁡−1(tan⁡(π4−x))- {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} - x} \right)} \right)

⇒\Rightarrow −(π4−x)- \left( {{\pi \over 4} - x} \right)

⇒\Rightarrow dydx=1{{dy} \over {dx}} = 1

Now if differentiation of x2{x \over 2} w.r.t is 12{1 \over 2}

⇒\Rightarrow So differentiation of y w.r.t x2{x \over 2} is 1121 \over {1 \over 2} = 2

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