If ` y=cos ^(-1) ((sqrt (1+x) -sqrt (1-x) )/( 2)) ,then (dy)/(dx)=`

Options

A. ` (1)/(2sqrt ( 1-x^(2)))`
B. ` (-1)/(2sqrt ( 1-x^(2)))`
C. ` (2)/(2sqrt ( 1-x^(2)))`
D. ` (-2)/(2sqrt ( 1-x^(2)))`

Answer

To solve the problem \( y = \cos^{-1} \left( \frac{\sqrt{1+x} - \sqrt{1-x}}{2} \right) \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Simplify the expression inside the inverse cosine
Let \( z = \frac{\sqrt{1+x} - \sqrt{1-x}}{2} \). We can simplify \( z \) using the identity for sine in terms of half angles.

### Step 2: Substitute \( x \) with \( \sin \theta \)
To simplify further, we can set \( x = \sin \theta \). This gives us:
\[ \sqrt{1+x} = \sqrt{1+\sin \theta} = \sqrt{2 \cos^2 \frac{\theta}{2}} = \sqrt{2} \cos \frac{\theta}{2} \]
\[ \sqrt{1-x} = \sqrt{1-\sin \theta} = \sqrt{2 \cos^2 \frac{\theta}{2}} = \sqrt{2} \sin \frac{\theta}{2} \]
Thus,
\[ z = \frac{\sqrt{2} \cos \frac{\theta}{2} - \sqrt{2} \sin \frac{\theta}{2}}{2} = \frac{\sqrt{2}}{2} \left( \cos \frac{\theta}{2} - \sin \frac{\theta}{2} \right) \]

### Step 3: Rewrite \( y \)
Now we can express \( y \) as:
\[ y = \cos^{-1} \left( \frac{\sqrt{2}}{2} \left( \cos \frac{\theta}{2} - \sin \frac{\theta}{2} \right) \right) \]
Using the identity \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \), we can express \( y \) as:
\[ y = \pi - \frac{\theta}{2} \]

### Step 4: Substitute back for \( \theta \)
Since \( \theta = \sin^{-1}(x) \), we have:
\[ y = \pi - \frac{1}{2} \sin^{-1}(x) \]

### Step 5: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 0 - \frac{1}{2} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \frac{d}{dx}(x) = -\frac{1}{2\sqrt{1-x^2}} \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{1-x^2}} \]