What is `(dsqrt(1-sin2x))/dx` equal to, where `pi/4ltxltpi/2` ?

Options

A. `cosx+sinx`
B. `-(cosx+sinx)`
C. `pm(cosx+sinx)`
D. None of the above

Answer

To find the derivative of the function \( \sqrt{1 - \sin(2x)} \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the function
The function can be expressed as:
\[ y = \sqrt{1 - \sin(2x)} \]

### Step 2: Apply the chain rule
To differentiate \( y \), we will use the chain rule. The derivative of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \). Here, \( u = 1 - \sin(2x) \).

### Step 3: Differentiate the inner function
Now we need to find \( \frac{du}{dx} \):
\[ u = 1 - \sin(2x) \]
Using the chain rule again, we differentiate \( \sin(2x) \):
\[ \frac{du}{dx} = -\cos(2x) \cdot \frac{d(2x)}{dx} = -2\cos(2x) \]

### Step 4: Combine the derivatives
Now we can substitute back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{1 - \sin(2x)}} \cdot (-2\cos(2x)) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\cos(2x)}{\sqrt{1 - \sin(2x)}} \]

### Step 5: Simplify further using trigonometric identities
We know that \( \sin(2x) = 2\sin(x)\cos(x) \). Therefore, we can rewrite \( 1 - \sin(2x) \) as:
\[ 1 - \sin(2x) = 1 - 2\sin(x)\cos(x) = (\sin^2(x) + \cos^2(x)) - 2\sin(x)\cos(x) = (\sin(x) - \cos(x))^2 \]
Thus,
\[ \sqrt{1 - \sin(2x)} = |\sin(x) - \cos(x)| \]

### Step 6: Determine the sign of \( \sin(x) - \cos(x) \)
Since we are given that \( \frac{\pi}{4} < x < \frac{\pi}{2} \), in this interval, \( \sin(x) > \cos(x) \). Therefore, we can drop the absolute value:
\[ \sqrt{1 - \sin(2x)} = \sin(x) - \cos(x) \]

### Step 7: Final derivative
Now substituting this back into our derivative:
\[ \frac{dy}{dx} = -\frac{\cos(2x)}{\sin(x) - \cos(x)} \]
Using the identity \( \cos(2x) = \cos^2(x) - \sin^2(x) \), we can express the final answer as:
\[ \frac{dy}{dx} = \cos(x) + \sin(x) \]

### Conclusion
Thus, the derivative \( \frac{d}{dx} \sqrt{1 - \sin(2x)} \) is:
\[ \frac{d}{dx} \sqrt{1 - \sin(2x)} = \cos(x) + \sin(x) \]