If `y=sin^(-1){(sqrt(1+x)-sqrt(1-x))/(2)}`, find `(dy)/(dx).`

Options

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

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)\), we will use the chain rule and implicit differentiation. Let's go through the steps:

### Step 1: Define the function
We start with the function:
\[ y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right) \]

### Step 2: Differentiate both sides
To find \(\frac{dy}{dx}\), we differentiate both sides with respect to \(x\):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)^2}} \cdot \frac{d}{dx}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right) \]

### Step 3: Differentiate the inner function
Now we need to differentiate the inner function:
\[ \frac{d}{dx}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right) = \frac{1}{2}\left(\frac{d}{dx}(\sqrt{1+x}) - \frac{d}{dx}(\sqrt{1-x})\right) \]
Using the chain rule:
\[ \frac{d}{dx}(\sqrt{1+x}) = \frac{1}{2\sqrt{1+x}} \quad \text{and} \quad \frac{d}{dx}(\sqrt{1-x}) = -\frac{1}{2\sqrt{1-x}} \]
Thus,
\[ \frac{d}{dx}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right) = \frac{1}{2}\left(\frac{1}{2\sqrt{1+x}} + \frac{1}{2\sqrt{1-x}}\right) \]
This simplifies to:
\[ \frac{1}{4}\left(\frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1-x}}\right) \]

### Step 4: Substitute back into the derivative
Now substituting back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)^2}} \cdot \frac{1}{4}\left(\frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1-x}}\right) \]

### Step 5: Simplify the expression
Next, we need to simplify \(\sqrt{1 - \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)^2}\):
\[ \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)^2 = \frac{(1+x) + (1-x) - 2\sqrt{(1+x)(1-x)}}{4} = \frac{2 - 2\sqrt{(1+x)(1-x)}}{4} = \frac{1 - \sqrt{(1+x)(1-x)}}{2} \]
Thus,
\[ 1 - \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)^2 = 1 - \frac{1 - \sqrt{(1+x)(1-x)}}{2} = \frac{1 + \sqrt{(1+x)(1-x)}}{2} \]
So,
\[ \sqrt{1 - \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)^2} = \sqrt{\frac{1 + \sqrt{(1+x)(1-x)}}{2}} \]

### Final expression for \(\frac{dy}{dx}\)
Putting it all together, we have:
\[ \frac{dy}{dx} = \frac{1}{4\sqrt{\frac{1 + \sqrt{(1+x)(1-x)}}{2}}} \left(\frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1-x}}\right) \]