Question

If \[ y=\tan^{-1}\left( \frac{\sqrt{1+x^3}+\sqrt{1-x^3}} {\sqrt{1+x^3}-\sqrt{1-x^3}} \right) \] then \(\dfrac{dy}{dx}\) is equal to:

• A. \(-\dfrac{3x^2}{2\sqrt{1-x^6}}\)
• B. \(\dfrac{6x^2}{\sqrt{1-x^6}}\)
• C. \(-\dfrac{6x^2}{\sqrt{1-x^6}}\)
• D. \(\dfrac{3x^2}{\sqrt{1-x^6}}\)

Hint:

For complicated inverse trigonometric expressions, try converting radicals into trigonometric substitutions. The expression often collapses into a standard identity.

Answer 1

The Correct Option is A

Concept: Expressions involving inverse tangent and radicals often simplify beautifully through algebraic manipulation and trigonometric identities. We use the identity: \[ \tan^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right) = \frac{\pi}{4}+\frac{\theta}{2} \] after simplifying the given expression.

Step 1: Simplify the expression inside the inverse tangent. Let \[ a=\sqrt{1+x^3}, \qquad b=\sqrt{1-x^3} \] Then, \[ y=\tan^{-1}\left(\frac{a+b}{a-b}\right) \] Multiply numerator and denominator by \(a+b\): \[ \frac{a+b}{a-b}\cdot\frac{a+b}{a+b} = \frac{(a+b)^2}{a^2-b^2} \] Now, \[ a^2-b^2=(1+x^3)-(1-x^3)=2x^3 \] and \[ (a+b)^2 = 1+x^3+1-x^3+2\sqrt{(1+x^3)(1-x^3)} \] \[ =2+2\sqrt{1-x^6} \] Thus, \[ \frac{a+b}{a-b} = \frac{2+2\sqrt{1-x^6}}{2x^3} \] \[ = \frac{1+\sqrt{1-x^6}}{x^3} \] Hence, \[ y=\tan^{-1}\left( \frac{1+\sqrt{1-x^6}}{x^3} \right) \]

Step 2: Use trigonometric substitution. Let \[ x^3=\sin\theta \] Then, \[ \sqrt{1-x^6}=\cos\theta \] Hence, \[ y=\tan^{-1}\left( \frac{1+\cos\theta}{\sin\theta} \right) \] Using identity, \[ \frac{1+\cos\theta}{\sin\theta} = \cot\frac{\theta}{2} \] Therefore, \[ y=\tan^{-1}\left(\cot\frac{\theta}{2}\right) \] \[ y=\frac{\pi}{2}-\frac{\theta}{2} \]

Step 3: Differentiate both sides. \[ \frac{dy}{dx} = -\frac12\frac{d\theta}{dx} \] Since, \[ x^3=\sin\theta \] differentiate: \[ 3x^2=\cos\theta\frac{d\theta}{dx} \] \[ \frac{d\theta}{dx} = \frac{3x^2}{\cos\theta} \] But, \[ \cos\theta=\sqrt{1-x^6} \] Hence, \[ \frac{d\theta}{dx} = \frac{3x^2}{\sqrt{1-x^6}} \] Therefore, \[ \frac{dy}{dx} = -\frac12\left( \frac{3x^2}{\sqrt{1-x^6}} \right) \] \[ \boxed{ \frac{dy}{dx} = -\frac{3x^2}{2\sqrt{1-x^6}} } \]