The derivative of `sin^(-1)((sqrt(1+x)+sqrt(1-x))/(2))` with respect to x is

Options

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

Answer

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

### Step 1: Simplify the Expression
Let:
\[ y = \sin^{-1}\left(\frac{\sqrt{1+x} + \sqrt{1-x}}{2}\right) \]
We can rewrite the expression inside the sine inverse function.

### Step 2: Use Trigonometric Identities
Let \( x = \cos(2\theta) \). Then:
\[ \sqrt{1+x} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2(\theta)} = \sqrt{2}\cos(\theta) \]
\[ \sqrt{1-x} = \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2(\theta)} = \sqrt{2}\sin(\theta) \]

### Step 3: Substitute Back
Substituting these back into our expression gives:
\[ y = \sin^{-1}\left(\frac{\sqrt{2}\cos(\theta) + \sqrt{2}\sin(\theta)}{2}\right) = \sin^{-1}\left(\frac{\sqrt{2}}{2}(\cos(\theta) + \sin(\theta))\right) \]
Since \( \frac{\sqrt{2}}{2} = \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \), we can write:
\[ y = \sin^{-1}\left(\sin\left(\frac{\pi}{4} + \theta\right)\right) \]
Thus:
\[ y = \frac{\pi}{4} + \theta \]

### Step 4: Find \( \theta \)
From \( x = \cos(2\theta) \), we have:
\[ 2\theta = \cos^{-1}(x) \quad \Rightarrow \quad \theta = \frac{1}{2}\cos^{-1}(x) \]
Substituting this back gives:
\[ y = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}(x) \]

### Step 5: Differentiate with Respect to \( x \)
Now, we differentiate \( y \):
\[ \frac{dy}{dx} = 0 + \frac{1}{2} \cdot \frac{d}{dx}(\cos^{-1}(x)) \]
The derivative of \( \cos^{-1}(x) \) is:
\[ \frac{d}{dx}(\cos^{-1}(x)) = -\frac{1}{\sqrt{1-x^2}} \]
Thus:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) = -\frac{1}{2\sqrt{1-x^2}} \]

### Final Answer
The derivative of \( \sin^{-1}\left(\frac{\sqrt{1+x} + \sqrt{1-x}}{2}\right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{1-x^2}} \]