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

Step 1: Simplify the expression inside the inverse sine. Using the identity \[ 1-\cos2x=2\sin^2x, \] we get \[ \frac{1-\cos2x}{1+\sin^4x} = \frac{2\sin^2x}{1+\sin^4x}. \] Again, \[ 1+\sin^4x = (\sin^2x)^2+1. \] Using the identity \[ \sin2\theta = \frac{2\tan\theta}{1+\tan^2\theta}, \] with \[ \tan\theta=\sin^2x, \] we obtain \[ \frac{2\sin^2x}{1+\sin^4x} = \sin\!\left(2\tan^{-1}(\sin^2x)\right). \] Hence, \[ y = 2\tan^{-1}(\sin^2x). \]

Step 2: Differentiate. Differentiating, \[ \frac{dy}{dx} = 2\cdot \frac{1}{1+\sin^4x} \cdot \frac{d}{dx}(\sin^2x). \] Since \[ \frac{d}{dx}(\sin^2x) = 2\sin x\cos x = \sin2x, \] we obtain \[ \boxed{ \frac{dy}{dx} = \frac{2\sin2x}{1+\sin^4x}. } \] Hence, the correct option is \(\boxed{(B)}\).