Question

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

• A. $2 \sec x \tan x$
• B. $\sin 2x$
• C. $2 \sec 2x \tan 2x$
• D. $\cos 2x$

Answer 1

Answer: (C)

$2 \sec 2x \tan 2x$

Answer 2

Begin by rewriting the expression for $y$ in terms of trigonometric identities:

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

Using the double angle identity $\sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x)$, rewrite $y$:

$y = \frac{1}{\sqrt{1 - \sin^2(2x)}}$.

The term $1 - \sin^2(2x)$ simplifies to $\cos^2(2x)$. Thus:

$y = \frac{1}{\cos(2x)} = \sec(2x)$.

Differentiate $y$ with respect to $x$:

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

The derivative of $\sec(2x)$ is:

$\frac{dy}{dx} = \sec(2x) \tan(2x) \times 2$.

Thus:

$\frac{dy}{dx} = 2 \sec^2(2x) \tan(2x)$.

Therefore, the correct answer is:

$2 \sec 2x \tan 2x$.