If $y =\frac{1}{\sqrt{1- 4 \sin^2x \cos^2x}}$, then $\frac{dy}{dx}=$
Answer & explanation
Correct answer: option 3
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:
$= \sec 2x \tan 2x \cdot 2$