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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

If y = ^ - 1 ( x - x), then the absolute value of 2 (dy)/(dx) is

If y=tan⁡−1(sec⁡x−tan⁡x),thentheabsolutevalueof2y = \tan^{- 1}\left(\sec x - \tan x\right), then the absolute value of 2\dfrac{dy}{dx}$ is

Correct answer

1

Step-by-step solution

y=tan⁡−1(sec⁡x−tan⁡x)y = \tan^{- 1}\left(\sec x - \tan x\right)
y=tan⁡−1(1−sin⁡xcos⁡x)y = \tan^{- 1}\left(\dfrac{1 - \sin x}{\cos x}\right)
⇒y=tan⁡−1((cosx2−sinx2)2(cos⁡2x2−sin⁡2x2))⇒ y = \tan^{- 1}\left(\dfrac{{\left(cos\dfrac{x}{2} - sin\dfrac{x}{2}\right)}^{2}}{\left(\cos^{2}\dfrac{x}{2} - \sin^{2}\dfrac{x}{2}\right)}\right)
⇒y=tan⁡−1(cosx2−sinx2cosx2+sinx2)⇒ y = \tan^{- 1}\left(\dfrac{cos\dfrac{x}{2} - sin\dfrac{x}{2}}{cos\dfrac{x}{2} + sin\dfrac{x}{2}}\right)
⇒y=tan⁡−1(1−tanx21+tanx2)⇒ y = \tan^{- 1}\left(\dfrac{1 - tan\dfrac{x}{2}}{1 + tan\dfrac{x}{2}}\right)
⇒y=tan⁡−1(tan⁡(π4−x2))⇒ y = \tan^{- 1}\left(\tan \left(\dfrac{\pi}{4} - \dfrac{x}{2}\right)\right)
⇒y=π4−x2⇒ y = \dfrac{\pi}{4} - \dfrac{x}{2}
Differentiating with respect to x,x, we get
⇒dydx=−12=−0.5⇒ \dfrac{dy}{dx} = \dfrac{- 1}{2} = - 0.5
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