Differentiate the following w.r.t. x :
`sin^(-1)(sqrt((1+x^(2))/2))`.

Answer

To differentiate the function \( y = \sin^{-1}\left(\sqrt{\frac{1+x^2}{2}}\right) \) with respect to \( x \), we will apply the chain rule and the derivative of the inverse sine function.

### Step-by-Step Solution:

1. **Identify the outer and inner functions**:
- Let \( u = \sqrt{\frac{1+x^2}{2}} \).
- Then, \( y = \sin^{-1}(u) \).

2. **Differentiate the outer function**:
- The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} \]

3. **Differentiate the inner function**:
- Now, we need to differentiate \( u \) with respect to \( x \):
\[ u = \sqrt{\frac{1+x^2}{2}} \implies u^2 = \frac{1+x^2}{2} \]
- Differentiate \( u^2 \) with respect to \( x \):
\[ \frac{d(u^2)}{dx} = \frac{d}{dx}\left(\frac{1+x^2}{2}\right) = \frac{1}{2} \cdot 2x = x \]
- Now, using the chain rule:
\[ \frac{du}{dx} = \frac{1}{2\sqrt{\frac{1+x^2}{2}}} \cdot \frac{d(u^2)}{dx} = \frac{1}{2\sqrt{\frac{1+x^2}{2}}} \cdot x = \frac{x}{2\sqrt{\frac{1+x^2}{2}}} \]

4. **Combine the derivatives using the chain rule**:
- Now, we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{x}{2\sqrt{\frac{1+x^2}{2}}} \]

5. **Substitute \( u \) back into the equation**:
- We know \( u = \sqrt{\frac{1+x^2}{2}} \), so we need to find \( 1-u^2 \):
\[ 1 - u^2 = 1 - \frac{1+x^2}{2} = \frac{2 - (1+x^2)}{2} = \frac{1 - x^2}{2} \]
- Therefore, \( \sqrt{1-u^2} = \sqrt{\frac{1-x^2}{2}} \).

6. **Final expression for \( \frac{dy}{dx} \)**:
- Substitute everything back:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{\frac{1-x^2}{2}}} \cdot \frac{x}{2\sqrt{\frac{1+x^2}{2}}} \]
- Simplifying gives:
\[ \frac{dy}{dx} = \frac{x}{2} \cdot \frac{\sqrt{2}}{\sqrt{1-x^2} \cdot \sqrt{1+x^2}} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{x \sqrt{2}}{2 \sqrt{(1-x^2)(1+x^2)}} \]