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

Question

If $ y = \log_e \left[ e^{3x} \left( \frac{x - 4}{x + 3} \right)^{3/2} \right] $, then find $ \frac{dy}{dx} $:

Answer 1

Step 1: Half angle trick
Write $\sqrt{\dfrac{1 + \sin 2x}{1 - \sin 2x}} = \dfrac{1 + \tan x}{1 - \tan x}$ after dividing by $\cos x$.

Step 2: Recognise
This is $\tan(\pi/4 + x)$, and $\pi/4 + \pi/6 = 5\pi/12$ is in the principal range.

Step 3: Differentiate
$y = \pi/4 + x$ so $y' = 1$. Option (B).

Final Answer:
Option (B). \[ \boxed{1} \]

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

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The Correct Option is B

Solution and Explanation

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