Question

If \[ y=\sin^{-1}\!\left(\frac{1-\cos2x}{1+\sin^{4}x}\right), \] then \[ \frac{dy}{dx}= \]

• A. \(\dfrac{2\cos2x}{1+\sin^{4}x}\)
• B. \(\dfrac{2\sin2x}{1+\sin^{4}x}\)
• C. \(\dfrac{2\cos2x}{1+\sin^{8}x}\)
• D. \(\dfrac{2\sin2x}{1+\sin^{8}x}\)

Hint:

Remember the useful identity \[ \boxed{ \sin\!\left(2\tan^{-1}t\right) = \frac{2t}{1+t^2}. } \] Whenever an expression is of the form \[ \frac{2t}{1+t^2}, \] replace it by \[ \sin\!\left(2\tan^{-1}t\right) \] to simplify inverse trigonometric differentiation.

Answer 1

The Correct Option is B