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

Options

A. ` (-x)/( sqrt(1-x^(4))) `
B. ` (x)/( sqrt(1-x^(4))) `
C. ` (-2x)/( sqrt(1-x^(4))) `
D. ` (2x)/( sqrt(1-x^(4))) `

Answer

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

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

### Step 2: Use the identity for sine inverse
We can express the square root in terms of cosine:
\[ \sqrt{\frac{1 + x^2}{2}} = \cos\left(\frac{\theta}{2}\right) \quad \text{where } \theta = \cos^{-1}(x) \]
Thus, we can rewrite \( y \) as:
\[ y = \sin^{-1} \left( \cos\left(\frac{\theta}{2}\right) \right) \]

### Step 3: Apply the sine inverse identity
Using the identity \( \sin^{-1}(\cos(\theta)) = \frac{\pi}{2} - \theta \):
\[ y = \frac{\pi}{2} - \frac{1}{2} \cos^{-1}(x^2) \]

### Step 4: Differentiate the function
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 0 - \frac{1}{2} \cdot \frac{d}{dx} \left( \cos^{-1}(x^2) \right) \]

### Step 5: Use the chain rule for differentiation
Using the chain rule:
\[ \frac{d}{dx} \left( \cos^{-1}(u) \right) = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
where \( u = x^2 \). Thus, we have:
\[ \frac{du}{dx} = 2x \]
So,
\[ \frac{d}{dx} \left( \cos^{-1}(x^2) \right) = -\frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x = -\frac{2x}{\sqrt{1 - x^4}} \]

### Step 6: Substitute back into the derivative
Substituting this back into our derivative expression:
\[ \frac{dy}{dx} = -\frac{1}{2} \left( -\frac{2x}{\sqrt{1 - x^4}} \right) = \frac{x}{\sqrt{1 - x^4}} \]

### Final Result
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{1 - x^4}} \]

---