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Differentiation question

2019 · 12 Apr · Shift 2 · Q34
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  5. /2019 · 12 Apr · Shift 2 · Q34

Differentiation question

2019 · 12 Apr · Shift 2 · Q34

JEE MainMathematicsDifferentiationMCQ+4 / −1
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)tan−1(sinx+cosxsinx−cosx​), with respect to x2{x \over 2}2x​, where (x∈(0,π2))\left( {x \in \left( {0,{\pi \over 2}} \right)} \right)(x∈(0,2π​)) is :
  1. A
    1
  2. B
    2
  3. C
    23{2 \over 3}32​
  4. D
    12{1 \over 2}21​
View written solutionFree

Correct answer: B

  1. Given function

We need the derivative of

y=tan⁡−1(sin⁡x−cos⁡xsin⁡x+cos⁡x)y=\tan^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)y=tan−1(sinx+cosxsinx−cosx​)

with respect to x2\dfrac{x}{2}2x​, where x∈(0,π2)x\in\left(0,\dfrac\pi2\right)x∈(0,2π​).

  1. Simplify the expression inside

Divide numerator and denominator by cos⁡x\cos xcosx:

sin⁡x−cos⁡xsin⁡x+cos⁡x=tan⁡x−1tan⁡x+1\frac{\sin x-\cos x}{\sin x+\cos x} =\frac{\tan x-1}{\tan x+1}sinx+cosxsinx−cosx​=tanx+1tanx−1​

Now use the identity

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}tan(A−B)=1+tanAtanBtanA−tanB​

So,

tan⁡x−11+tan⁡x=tan⁡(x−π4)\frac{\tan x-1}{1+\tan x} = \tan\left(x-\frac\pi4\right)1+tanxtanx−1​=tan(x−4π​)

Hence,

y=tan⁡−1(tan⁡(x−π4))y=\tan^{-1}\left(\tan\left(x-\frac\pi4\right)\right)y=tan−1(tan(x−4π​))
  1. Use the interval condition

Since

x∈(0,π2)x\in\left(0,\frac\pi2\right)x∈(0,2π​)

we have

x−π4∈(−π4,π4)x-\frac\pi4\in\left(-\frac\pi4,\frac\pi4\right)x−4π​∈(−4π​,4π​)

This lies completely inside the principal range of tan⁡−1\tan^{-1}tan−1, i.e.

(−π2,π2)\left(-\frac\pi2,\frac\pi2\right)(−2π​,2π​)

Therefore,

tan⁡−1(tan⁡θ)=θ\tan^{-1}(\tan\theta)=\thetatan−1(tanθ)=θ

for θ=x−π4\theta=x-\frac\pi4θ=x−4π​.

So,

y=x−π4y=x-\frac\pi4y=x−4π​
  1. Differentiate with respect to x/2x/2x/2

Let

u=x2u=\frac{x}{2}u=2x​

Then x=2ux=2ux=2u, so

dydu=dydx⋅dxdu\frac{dy}{du}=\frac{dy}{dx}\cdot\frac{dx}{du}dudy​=dxdy​⋅dudx​

Now,

dydx=ddx(x−π4)=1\frac{dy}{dx}=\frac{d}{dx}\left(x-\frac\pi4\right)=1dxdy​=dxd​(x−4π​)=1

and

dxdu=2\frac{dx}{du}=2dudx​=2

Thus,

dyd(x/2)=1⋅2=2\frac{dy}{d(x/2)}=1\cdot 2=2d(x/2)dy​=1⋅2=2
  1. Check options

The correct option is:

2\boxed{2}2​

So, Option B is correct.

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