If $y =\frac{1}{\sqrt{1- 4 \sin^2x \cos^2x}}$, then $\frac{dy}{dx}=$

$2 \sec 2x \tan 2x$

The correct answer is Option (3) → $2 \sec 2x \tan 2x$

We are given:

$y = \frac{1}{\sqrt{1 - 4 \sin^2 x \cos^2 x}}$

Step 1: Simplify the expression

Use identity:

$\sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x$

So,

$4 \sin^2 x \cos^2 x = \sin^2 2x$

Thus,

$y = \frac{1}{\sqrt{1 - \sin^2 2x}} = \frac{1}{\sqrt{\cos^2 2x}} = \frac{1}{|\cos 2x|}$

For standard domain, take:

$y = \sec 2x$

Step 2: Differentiate

$\frac{dy}{dx} = \frac{d}{dx}(\sec 2x)$

Using chain rule:

$= \sec 2x \tan 2x \cdot 2$