`(d)/(dx)[sqrt((1-sin2x)/(1+sin2x))]=`

Options

A. `sec^(2)x`
B. `-sec^(2)((pi)/(4)-x)`
C. `sec^(2)((pi)/(4)-x)`
D. none of theses

Answer

To find the derivative of the function \( y = \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} \), we will follow these steps:

### Step 1: Simplify the expression inside the square root
We start with the expression:
\[ y = \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} \]
Using the half-angle identity, we can express \( \sin 2x \) as \( 2 \sin x \cos x \):
\[ y = \sqrt{\frac{1 - 2 \sin x \cos x}{1 + 2 \sin x \cos x}} \]

### Step 2: Rewrite the numerator and denominator
We can rewrite \( 1 \) in both the numerator and the denominator as \( \cos^2 x + \sin^2 x \):
\[ y = \sqrt{\frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x + \sin^2 x + 2 \sin x \cos x}} = \sqrt{\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}} \]

### Step 3: Simplify the square root
Now we can simplify the square root:
\[ y = \frac{\cos x - \sin x}{\cos x + \sin x} \]

### Step 4: Differentiate using the quotient rule
We will differentiate \( y \) using the quotient rule:
\[ \frac{dy}{dx} = \frac{(u'v - uv')}{v^2} \]
where \( u = \cos x - \sin x \) and \( v = \cos x + \sin x \).

Calculating \( u' \) and \( v' \):
- \( u' = -\sin x - \cos x \)
- \( v' = -\sin x + \cos x \)

Now substituting into the quotient rule:
\[ \frac{dy}{dx} = \frac{(-\sin x - \cos x)(\cos x + \sin x) - (\cos x - \sin x)(-\sin x + \cos x)}{(\cos x + \sin x)^2} \]

### Step 5: Simplify the derivative
Expanding the numerator:
1. First term:
\[ (-\sin x - \cos x)(\cos x + \sin x) = -\sin x \cos x - \sin^2 x - \cos^2 x - \cos x \sin x = -2\sin x \cos x - 1 \]
2. Second term:
\[ (\cos x - \sin x)(-\sin x + \cos x) = -\cos x \sin x + \cos^2 x + \sin^2 x - \sin x \cos x = 1 - 2\sin x \cos x \]

Combining these:
\[ \frac{dy}{dx} = \frac{-2\sin x \cos x - 1 - (1 - 2\sin x \cos x)}{(\cos x + \sin x)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-2}{(\cos x + \sin x)^2} \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = -\frac{2}{(\cos x + \sin x)^2} \]

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