Question

If \(y = tan^{-1}\sqrt{\frac{1+sin2x}{1-sin2x}}\), then \(\frac{\text{d}y}{\text{d}x}\) at \(x = \frac{π}{6}\) is

• A. \(0\)
• B. \(1\)
• C. \(\frac{1}{2}\)
• D. \(\frac{\sqrt{3}}{2}\)

Hint:

Simplify the surd to tan(pi/4 + x) before differentiating.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Use \(1 + \sin2x = (\cos x + \sin x)^2\) and \(1 - \sin2x = (\cos x - \sin x)^2\).

Step 2: Simplify
\[ \sqrt{\frac{1 + \sin2x}{1 - \sin2x}} = \frac{\cos x + \sin x}{\cos x - \sin x} = \frac{1 + \tan x}{1 - \tan x} = \tan\left(\frac{\pi}{4} + x\right) \]
For \(x\) near \(\pi/6\), \(\frac\pi4 + x\) lies in \((-\frac\pi2, \frac\pi2)\), so \(y = \tan^{-1}\tan(\frac\pi4 + x) = \frac\pi4 + x\).

Step 3: Differentiate
\[ \frac{dy}{dx} = 1 \]
The derivative is 1 for all x in that range, including \(x = \pi/6\).

Final Answer:
The derivative at \(x = \frac\pi6\) is 1, option (B). \[ \boxed{1} \]