If `f(x)=sqrt(1-sin2x),` then `f'(x)` equals

Options

A. `-(cosx+sinx)." for "x in(pi//4,pi//2)`
B. `cosx+sinx," fro "x in(0,pi//4)`
C. `-(cosx+sinx)," for "x in(0,pi//4)`
D. `cosx-sinx," for "x in(pi//4,pi//2)`

Answer

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

### Step 1: Rewrite the function using trigonometric identities
We know that \( \sin(2x) = 2\sin(x)\cos(x) \). Therefore, we can rewrite the function as:
\[ f(x) = \sqrt{1 - 2\sin(x)\cos(x)} \]

### Step 2: Differentiate using the chain rule
To differentiate \( f(x) \), we will use the chain rule. The derivative of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \cdot u' \). Here, \( u = 1 - 2\sin(x)\cos(x) \).

First, we need to find \( u' \):
\[ u = 1 - 2\sin(x)\cos(x) \]
Using the product rule, we differentiate \( -2\sin(x)\cos(x) \):
\[ u' = -2(\cos(x)\cdot\cos(x) + \sin(x)(-\sin(x))) = -2(\cos^2(x) - \sin^2(x)) = -2\cos^2(x) + 2\sin^2(x) \]

### Step 3: Apply the chain rule
Now we can apply the chain rule:
\[ f'(x) = \frac{1}{2\sqrt{1 - 2\sin(x)\cos(x)}} \cdot (-2\cos^2(x) + 2\sin^2(x)) \]
This simplifies to:
\[ f'(x) = \frac{-\cos^2(x) + \sin^2(x)}{\sqrt{1 - 2\sin(x)\cos(x)}} \]

### Step 4: Simplify the expression
We can further simplify:
\[ f'(x) = \frac{\sin^2(x) - \cos^2(x)}{\sqrt{1 - 2\sin(x)\cos(x)}} \]

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = \frac{\sin^2(x) - \cos^2(x)}{\sqrt{1 - 2\sin(x)\cos(x)}} \]

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