Q81 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics

If $y = \tan^{-1}\left(\dfrac{\sqrt{1+x^3} + \sqrt{1-x^3}}{\sqrt{1+x^3} - \sqrt{1-x^3}}\right)$ then $\dfrac{dy}{dx} =$

  • A
    $\dfrac{6x^2}{\sqrt{1-x^6}}$
  • B
    $-\dfrac{6x^2}{\sqrt{1-x^6}}$
  • C
    $\dfrac{3x^2}{\sqrt{1-x^6}}$
  • Correct answer
    D
    $-\dfrac{3x^2}{2\sqrt{1-x^6}}$

Solution

Substitute $x^3 = \cos 2\theta$.

Then $\sqrt{1+x^3} = \sqrt{1+\cos 2\theta} = \sqrt{2}\cos\theta$

And $\sqrt{1-x^3} = \sqrt{1-\cos 2\theta} = \sqrt{2}\sin\theta$

$y = \tan^{-1}\left(\dfrac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}\right)$

$y = \tan^{-1}\left(\dfrac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta}\right)$

Dividing the numerator and denominator by $\cos\theta$:

$y = \tan^{-1}\left(\dfrac{1 + \tan\theta}{1 - \tan\theta}\right)$

$y = \tan^{-1}\left(\tan\left(\dfrac{\pi}{4} + \theta\right)\right)$

$y = \dfrac{\pi}{4} + \theta$

Since $x^3 = \cos 2\theta$, we have $\theta = \dfrac{1}{2}\cos^{-1}(x^3)$.

$y = \dfrac{\pi}{4} + \dfrac{1}{2}\cos^{-1}(x^3)$

Differentiating with respect to $x$:

$\dfrac{dy}{dx} = \dfrac{1}{2} \left( -\dfrac{1}{\sqrt{1 - (x^3)^2}} \right) \cdot \dfrac{d}{dx}(x^3)$

$\dfrac{dy}{dx} = -\dfrac{3x^2}{2\sqrt{1 - x^6}}$

Answer: $-\dfrac{3x^2}{2\sqrt{1-x^6}}$

Hence, the correct answer is option: 'D'.

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