If `y=tan^(-1) sqrt((1-sin 4x)/(1+sin 4x)),"then " dy/dx=`

Options

A. ` 2`
B. `-2`
C. `4`
D. `-4`

Answer

To find the derivative of the function \( y = \tan^{-1} \left( \sqrt{\frac{1 - \sin 4x}{1 + \sin 4x}} \right) \), we will follow these steps:

### Step 1: Simplify the Expression Inside the Arctangent
We start with the expression inside the arctangent:
\[ y = \tan^{-1} \left( \sqrt{\frac{1 - \sin 4x}{1 + \sin 4x}} \right) \]

Using the identity for sine, we can rewrite \( \sin 4x \):
\[ \sin 4x = 2 \sin 2x \cos 2x \]
Thus, we can express \( 1 - \sin 4x \) and \( 1 + \sin 4x \) as:
\[ 1 - \sin 4x = 1 - 2 \sin 2x \cos 2x = \cos^2 2x + \sin^2 2x - 2 \sin 2x \cos 2x = (\cos 2x - \sin 2x)^2 \]
\[ 1 + \sin 4x = 1 + 2 \sin 2x \cos 2x = \cos^2 2x + \sin^2 2x + 2 \sin 2x \cos 2x = (\cos 2x + \sin 2x)^2 \]

Substituting these back into the expression gives:
\[ y = \tan^{-1} \left( \sqrt{\frac{(\cos 2x - \sin 2x)^2}{(\cos 2x + \sin 2x)^2}} \right) \]
This simplifies to:
\[ y = \tan^{-1} \left( \frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x} \right) \]

### Step 2: Use the Tangent Difference Identity
We can use the tangent subtraction formula:
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
Let \( a = \frac{\pi}{4} \) and \( b = 2x \):
\[ y = \tan^{-1} \left( \tan\left(\frac{\pi}{4} - 2x\right) \right) \]
Thus, we have:
\[ y = \frac{\pi}{4} - 2x \]

### Step 3: Differentiate
Now we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - 2x \right) \]
Since \( \frac{\pi}{4} \) is a constant, its derivative is \( 0 \), and the derivative of \( -2x \) is \( -2 \):
\[ \frac{dy}{dx} = 0 - 2 = -2 \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -2 \]

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