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